Relationship between prime factorizations of $n$ and $n+1$?












15














Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?



Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?










share|cite|improve this question
























  • Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
    – user17794
    Jul 20 '12 at 4:50






  • 2




    In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
    – Mark Bennet
    Jul 20 '12 at 4:53






  • 4




    $n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
    – Arturo Magidin
    Jul 20 '12 at 5:18
















15














Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?



Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?










share|cite|improve this question
























  • Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
    – user17794
    Jul 20 '12 at 4:50






  • 2




    In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
    – Mark Bennet
    Jul 20 '12 at 4:53






  • 4




    $n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
    – Arturo Magidin
    Jul 20 '12 at 5:18














15












15








15


10





Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?



Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?










share|cite|improve this question















Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?



Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?







number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 20 '12 at 4:42









Henry T. Horton

15k54463




15k54463










asked Jul 20 '12 at 4:40









EulerEuler

7613




7613












  • Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
    – user17794
    Jul 20 '12 at 4:50






  • 2




    In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
    – Mark Bennet
    Jul 20 '12 at 4:53






  • 4




    $n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
    – Arturo Magidin
    Jul 20 '12 at 5:18


















  • Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
    – user17794
    Jul 20 '12 at 4:50






  • 2




    In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
    – Mark Bennet
    Jul 20 '12 at 4:53






  • 4




    $n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
    – Arturo Magidin
    Jul 20 '12 at 5:18
















Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
– user17794
Jul 20 '12 at 4:50




Seems unlikely, given the extreme cases of Fermat and Mersenne primes.
– user17794
Jul 20 '12 at 4:50




2




2




In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
– Mark Bennet
Jul 20 '12 at 4:53




In general I think not (though the experts will confirm this), but if $n$ has two factors (not necessarily prime) which differ by 2 so that $n=r(r+2)$ then $n+1=(r+1)^2$.
– Mark Bennet
Jul 20 '12 at 4:53




4




4




$n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
– Arturo Magidin
Jul 20 '12 at 5:18




$n$ and $n+1$ are relatively prime, so no prime factor of one can be a prime factor of the other. But beyond that, there is no regularity; if $n$ is prime and different from $2$, then $n+1$ is certainly not prime; and conversely for $n+1neq 3$. There is no known regularity in the functions $omega(n)$, $Omega(n)$.
– Arturo Magidin
Jul 20 '12 at 5:18










6 Answers
6






active

oldest

votes


















9














Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.



Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture
and The Collatz Conjecture.






share|cite|improve this answer





























    9














    There is no known explicit relation between the prime factors of $n+1$ given the prime factorizaion of $n$.



    In fact, this is considered one of the hardest problems in our current understanding of number theory. Paul Erdős once famously quoted, albeit within the context of the Collatz conjecture - closely linked to the prime factorization of consecutive integers - that:



    "Mathematics is not yet ready for such problems".



    We can however deduce some basic properties for $n+1$ in the following way. If we denote by $omega(n)$ the number of distinct prime factors of $n$ and by $alpha_i$ the multiplicity of the $i^{th}$ prime in its decomposition, we have:



    $$
    n = prod_{i=1}^{omega(n)} p_i^{alpha_i}.
    $$



    From this we get:




    • $n+1$ is not divisible by any of the $p_i$, otherwise $p_i$ would divide $(n+1)-n$, which is clearly impossible because no prime number divides $1$,

    • If $n$ is even, $n+1$ is odd and vice-versa, which can trivially be extended to congruence modulo $p_i$,

    • There is no obvious relation between $omega(n)$ and $omega(n+1)$.


    One less obvious "observation" is Wilson's theorem. It states that if $p$ is a prime number, we have the following congruence relation:



    $$
    (p-1)! equiv -1 pmod p
    $$



    This connects the prime $p$ with its immediate integer predecessor $p-1$.



    There are also some non-trivial observations made by Erdős and Pomerance$^1$, which are the following. Define $P(n)$ as the largest prime factor of $n$. Then:




    • $P(n)>P(n+1), P(n+1)>P(n+2), P(n)<P(n+1)$ and $P(n+1)<P(n+2)$ occur infinitely often,

    • $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often,

    • $P(n)$ and $P(n+1)$ are usually not close, i.e., for each $epsilon>0$, there is a $delta>0$ such that for sufficiently large $x$, the number of $nleq x$ with


    $$
    x^{-delta}<P(n)/P(n+1)<x^{delta}
    $$



    is less than $epsilon x$.




    • Any integer $nleq x$ is divisible by at most $log(x)/log(2)$ primes.




    $^1$ Erdős and Pomerance, "On the largest prime factors of $n$ and $n+1$", Aequationes Mathematicae 17 (1978), available at: https://www.math.dartmouth.edu/~carlp/PDF/paper17.pdf






    share|cite|improve this answer



















    • 1




      You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
      – Malcolm
      Dec 7 '17 at 12:47












    • @Malcolm Noted!
      – Klangen
      Feb 6 '18 at 7:31










    • Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
      – idmercer
      Jun 13 '18 at 15:29



















    7














    While the factorisation of $N$ might not help much with the factorisation of $N+1$, in special circumstances, it can help with determining primality of $N$.



    Famous examples of this include Pépin's Test for primality of numbers of the form $2^{2^n}+1$, and Proth's Theorem for primality of numbers of the form $k times 2^n+1$ where $k<2^n$ (Proth primes feature on the top 20 known primes).



    There are more general primality tests for $N+1$ based on (partial) knowledge of the factorisation of $N$, but they tend to be less elegant. For example, this was snipped from "Factorizations of $b^n pm 1$, b = 2, 3, 5, 6, 7, 10, 11, 12 Up to High Powers" by Brillhart, Lehmer, Selfridge, Tuckerman, and Wagstaff, Jr.:




    Theorem 11. Let $N-1=FR$, where $F$ is completely factored and $(F,R)=1$. Suppose there exists an $a$ for which $a^{N-1} equiv 1 pmod N$ and $(a^{(N-1)/q}-1,N)=1$ for each prime factor $q$ of $F$. Let $R=rF+s$, $1 leq s < F$, and suppose $N<2F^3+2F$, $F>2$. If $r$ is odd, or if $r$ is even and $s^2-4r=t^2$, then $N$ is prime. Otherwise, $s^2-4r=t^2$ and $$N = [frac{1}{2}(s-t)F+1][frac{1}{2}(s+t)F+1].$$







    share|cite|improve this answer





























      6














      As I wrote when this question was raised at MathOverflow, if knowing the factorization of $n$ told you much about the factorization of $n+1$, the Fermat numbers $2^{2^n}+1$ would be easy --- but, they aren't.






      share|cite|improve this answer































        4














        Finding the connection of the prime factorization of n and n+1 would go hand in hand with proving Collatz Conjecture.






        share|cite|improve this answer

















        • 3




          Would it really?
          – Lord Shark the Unknown
          May 29 '17 at 20:52










        • I already heard that. Can you give details?
          – DaBler
          Nov 2 '18 at 19:35










        • DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
          – Collag3n
          Dec 5 '18 at 19:01



















        1














        I found a relation, here a proff:



        We know that:



        $n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



        $alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



        Then:



        $n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



        Where:



        $beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



        In other words:



        $n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



        $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



        Finally, this is the relation:



        $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



        In summary, Eq.1 can be used as a method for decomposition in prime factors, let's see an example:



        $n=60$



        $beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



        Then:



        $beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



        $beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



        $beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



        Finally:



        $60=2^{2}3^{1}5^{1}$






        share|cite|improve this answer























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          6 Answers
          6






          active

          oldest

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          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          9














          Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.



          Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture
          and The Collatz Conjecture.






          share|cite|improve this answer


























            9














            Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.



            Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture
            and The Collatz Conjecture.






            share|cite|improve this answer
























              9












              9








              9






              Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.



              Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture
              and The Collatz Conjecture.






              share|cite|improve this answer












              Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.



              Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture
              and The Collatz Conjecture.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jul 20 '12 at 5:16









              Ragib ZamanRagib Zaman

              28.5k34889




              28.5k34889























                  9














                  There is no known explicit relation between the prime factors of $n+1$ given the prime factorizaion of $n$.



                  In fact, this is considered one of the hardest problems in our current understanding of number theory. Paul Erdős once famously quoted, albeit within the context of the Collatz conjecture - closely linked to the prime factorization of consecutive integers - that:



                  "Mathematics is not yet ready for such problems".



                  We can however deduce some basic properties for $n+1$ in the following way. If we denote by $omega(n)$ the number of distinct prime factors of $n$ and by $alpha_i$ the multiplicity of the $i^{th}$ prime in its decomposition, we have:



                  $$
                  n = prod_{i=1}^{omega(n)} p_i^{alpha_i}.
                  $$



                  From this we get:




                  • $n+1$ is not divisible by any of the $p_i$, otherwise $p_i$ would divide $(n+1)-n$, which is clearly impossible because no prime number divides $1$,

                  • If $n$ is even, $n+1$ is odd and vice-versa, which can trivially be extended to congruence modulo $p_i$,

                  • There is no obvious relation between $omega(n)$ and $omega(n+1)$.


                  One less obvious "observation" is Wilson's theorem. It states that if $p$ is a prime number, we have the following congruence relation:



                  $$
                  (p-1)! equiv -1 pmod p
                  $$



                  This connects the prime $p$ with its immediate integer predecessor $p-1$.



                  There are also some non-trivial observations made by Erdős and Pomerance$^1$, which are the following. Define $P(n)$ as the largest prime factor of $n$. Then:




                  • $P(n)>P(n+1), P(n+1)>P(n+2), P(n)<P(n+1)$ and $P(n+1)<P(n+2)$ occur infinitely often,

                  • $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often,

                  • $P(n)$ and $P(n+1)$ are usually not close, i.e., for each $epsilon>0$, there is a $delta>0$ such that for sufficiently large $x$, the number of $nleq x$ with


                  $$
                  x^{-delta}<P(n)/P(n+1)<x^{delta}
                  $$



                  is less than $epsilon x$.




                  • Any integer $nleq x$ is divisible by at most $log(x)/log(2)$ primes.




                  $^1$ Erdős and Pomerance, "On the largest prime factors of $n$ and $n+1$", Aequationes Mathematicae 17 (1978), available at: https://www.math.dartmouth.edu/~carlp/PDF/paper17.pdf






                  share|cite|improve this answer



















                  • 1




                    You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                    – Malcolm
                    Dec 7 '17 at 12:47












                  • @Malcolm Noted!
                    – Klangen
                    Feb 6 '18 at 7:31










                  • Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                    – idmercer
                    Jun 13 '18 at 15:29
















                  9














                  There is no known explicit relation between the prime factors of $n+1$ given the prime factorizaion of $n$.



                  In fact, this is considered one of the hardest problems in our current understanding of number theory. Paul Erdős once famously quoted, albeit within the context of the Collatz conjecture - closely linked to the prime factorization of consecutive integers - that:



                  "Mathematics is not yet ready for such problems".



                  We can however deduce some basic properties for $n+1$ in the following way. If we denote by $omega(n)$ the number of distinct prime factors of $n$ and by $alpha_i$ the multiplicity of the $i^{th}$ prime in its decomposition, we have:



                  $$
                  n = prod_{i=1}^{omega(n)} p_i^{alpha_i}.
                  $$



                  From this we get:




                  • $n+1$ is not divisible by any of the $p_i$, otherwise $p_i$ would divide $(n+1)-n$, which is clearly impossible because no prime number divides $1$,

                  • If $n$ is even, $n+1$ is odd and vice-versa, which can trivially be extended to congruence modulo $p_i$,

                  • There is no obvious relation between $omega(n)$ and $omega(n+1)$.


                  One less obvious "observation" is Wilson's theorem. It states that if $p$ is a prime number, we have the following congruence relation:



                  $$
                  (p-1)! equiv -1 pmod p
                  $$



                  This connects the prime $p$ with its immediate integer predecessor $p-1$.



                  There are also some non-trivial observations made by Erdős and Pomerance$^1$, which are the following. Define $P(n)$ as the largest prime factor of $n$. Then:




                  • $P(n)>P(n+1), P(n+1)>P(n+2), P(n)<P(n+1)$ and $P(n+1)<P(n+2)$ occur infinitely often,

                  • $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often,

                  • $P(n)$ and $P(n+1)$ are usually not close, i.e., for each $epsilon>0$, there is a $delta>0$ such that for sufficiently large $x$, the number of $nleq x$ with


                  $$
                  x^{-delta}<P(n)/P(n+1)<x^{delta}
                  $$



                  is less than $epsilon x$.




                  • Any integer $nleq x$ is divisible by at most $log(x)/log(2)$ primes.




                  $^1$ Erdős and Pomerance, "On the largest prime factors of $n$ and $n+1$", Aequationes Mathematicae 17 (1978), available at: https://www.math.dartmouth.edu/~carlp/PDF/paper17.pdf






                  share|cite|improve this answer



















                  • 1




                    You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                    – Malcolm
                    Dec 7 '17 at 12:47












                  • @Malcolm Noted!
                    – Klangen
                    Feb 6 '18 at 7:31










                  • Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                    – idmercer
                    Jun 13 '18 at 15:29














                  9












                  9








                  9






                  There is no known explicit relation between the prime factors of $n+1$ given the prime factorizaion of $n$.



                  In fact, this is considered one of the hardest problems in our current understanding of number theory. Paul Erdős once famously quoted, albeit within the context of the Collatz conjecture - closely linked to the prime factorization of consecutive integers - that:



                  "Mathematics is not yet ready for such problems".



                  We can however deduce some basic properties for $n+1$ in the following way. If we denote by $omega(n)$ the number of distinct prime factors of $n$ and by $alpha_i$ the multiplicity of the $i^{th}$ prime in its decomposition, we have:



                  $$
                  n = prod_{i=1}^{omega(n)} p_i^{alpha_i}.
                  $$



                  From this we get:




                  • $n+1$ is not divisible by any of the $p_i$, otherwise $p_i$ would divide $(n+1)-n$, which is clearly impossible because no prime number divides $1$,

                  • If $n$ is even, $n+1$ is odd and vice-versa, which can trivially be extended to congruence modulo $p_i$,

                  • There is no obvious relation between $omega(n)$ and $omega(n+1)$.


                  One less obvious "observation" is Wilson's theorem. It states that if $p$ is a prime number, we have the following congruence relation:



                  $$
                  (p-1)! equiv -1 pmod p
                  $$



                  This connects the prime $p$ with its immediate integer predecessor $p-1$.



                  There are also some non-trivial observations made by Erdős and Pomerance$^1$, which are the following. Define $P(n)$ as the largest prime factor of $n$. Then:




                  • $P(n)>P(n+1), P(n+1)>P(n+2), P(n)<P(n+1)$ and $P(n+1)<P(n+2)$ occur infinitely often,

                  • $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often,

                  • $P(n)$ and $P(n+1)$ are usually not close, i.e., for each $epsilon>0$, there is a $delta>0$ such that for sufficiently large $x$, the number of $nleq x$ with


                  $$
                  x^{-delta}<P(n)/P(n+1)<x^{delta}
                  $$



                  is less than $epsilon x$.




                  • Any integer $nleq x$ is divisible by at most $log(x)/log(2)$ primes.




                  $^1$ Erdős and Pomerance, "On the largest prime factors of $n$ and $n+1$", Aequationes Mathematicae 17 (1978), available at: https://www.math.dartmouth.edu/~carlp/PDF/paper17.pdf






                  share|cite|improve this answer














                  There is no known explicit relation between the prime factors of $n+1$ given the prime factorizaion of $n$.



                  In fact, this is considered one of the hardest problems in our current understanding of number theory. Paul Erdős once famously quoted, albeit within the context of the Collatz conjecture - closely linked to the prime factorization of consecutive integers - that:



                  "Mathematics is not yet ready for such problems".



                  We can however deduce some basic properties for $n+1$ in the following way. If we denote by $omega(n)$ the number of distinct prime factors of $n$ and by $alpha_i$ the multiplicity of the $i^{th}$ prime in its decomposition, we have:



                  $$
                  n = prod_{i=1}^{omega(n)} p_i^{alpha_i}.
                  $$



                  From this we get:




                  • $n+1$ is not divisible by any of the $p_i$, otherwise $p_i$ would divide $(n+1)-n$, which is clearly impossible because no prime number divides $1$,

                  • If $n$ is even, $n+1$ is odd and vice-versa, which can trivially be extended to congruence modulo $p_i$,

                  • There is no obvious relation between $omega(n)$ and $omega(n+1)$.


                  One less obvious "observation" is Wilson's theorem. It states that if $p$ is a prime number, we have the following congruence relation:



                  $$
                  (p-1)! equiv -1 pmod p
                  $$



                  This connects the prime $p$ with its immediate integer predecessor $p-1$.



                  There are also some non-trivial observations made by Erdős and Pomerance$^1$, which are the following. Define $P(n)$ as the largest prime factor of $n$. Then:




                  • $P(n)>P(n+1), P(n+1)>P(n+2), P(n)<P(n+1)$ and $P(n+1)<P(n+2)$ occur infinitely often,

                  • $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often,

                  • $P(n)$ and $P(n+1)$ are usually not close, i.e., for each $epsilon>0$, there is a $delta>0$ such that for sufficiently large $x$, the number of $nleq x$ with


                  $$
                  x^{-delta}<P(n)/P(n+1)<x^{delta}
                  $$



                  is less than $epsilon x$.




                  • Any integer $nleq x$ is divisible by at most $log(x)/log(2)$ primes.




                  $^1$ Erdős and Pomerance, "On the largest prime factors of $n$ and $n+1$", Aequationes Mathematicae 17 (1978), available at: https://www.math.dartmouth.edu/~carlp/PDF/paper17.pdf







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 6 '18 at 7:31

























                  answered Sep 11 '17 at 11:12









                  KlangenKlangen

                  1,66211334




                  1,66211334








                  • 1




                    You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                    – Malcolm
                    Dec 7 '17 at 12:47












                  • @Malcolm Noted!
                    – Klangen
                    Feb 6 '18 at 7:31










                  • Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                    – idmercer
                    Jun 13 '18 at 15:29














                  • 1




                    You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                    – Malcolm
                    Dec 7 '17 at 12:47












                  • @Malcolm Noted!
                    – Klangen
                    Feb 6 '18 at 7:31










                  • Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                    – idmercer
                    Jun 13 '18 at 15:29








                  1




                  1




                  You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                  – Malcolm
                  Dec 7 '17 at 12:47






                  You attribute the last results to Erdős, but the paper you cite is by Erdős and Pomerance. I suggest that you either attribute the results to "Erdős and Pomerance" in your answer or find a reference authored by Erdős alone.
                  – Malcolm
                  Dec 7 '17 at 12:47














                  @Malcolm Noted!
                  – Klangen
                  Feb 6 '18 at 7:31




                  @Malcolm Noted!
                  – Klangen
                  Feb 6 '18 at 7:31












                  Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                  – idmercer
                  Jun 13 '18 at 15:29




                  Does the Erdos-Pomerance paper actually show that $P(n)>P(n+1)>P(n+2)$ does not occur infinitely often? My impression is that they're saying that they personally could not find a proof that $P(n)>P(n+1)>P(n+2)$ occurs infinitely often, but perhaps it does in fact occur infinitely often.
                  – idmercer
                  Jun 13 '18 at 15:29











                  7














                  While the factorisation of $N$ might not help much with the factorisation of $N+1$, in special circumstances, it can help with determining primality of $N$.



                  Famous examples of this include Pépin's Test for primality of numbers of the form $2^{2^n}+1$, and Proth's Theorem for primality of numbers of the form $k times 2^n+1$ where $k<2^n$ (Proth primes feature on the top 20 known primes).



                  There are more general primality tests for $N+1$ based on (partial) knowledge of the factorisation of $N$, but they tend to be less elegant. For example, this was snipped from "Factorizations of $b^n pm 1$, b = 2, 3, 5, 6, 7, 10, 11, 12 Up to High Powers" by Brillhart, Lehmer, Selfridge, Tuckerman, and Wagstaff, Jr.:




                  Theorem 11. Let $N-1=FR$, where $F$ is completely factored and $(F,R)=1$. Suppose there exists an $a$ for which $a^{N-1} equiv 1 pmod N$ and $(a^{(N-1)/q}-1,N)=1$ for each prime factor $q$ of $F$. Let $R=rF+s$, $1 leq s < F$, and suppose $N<2F^3+2F$, $F>2$. If $r$ is odd, or if $r$ is even and $s^2-4r=t^2$, then $N$ is prime. Otherwise, $s^2-4r=t^2$ and $$N = [frac{1}{2}(s-t)F+1][frac{1}{2}(s+t)F+1].$$







                  share|cite|improve this answer


























                    7














                    While the factorisation of $N$ might not help much with the factorisation of $N+1$, in special circumstances, it can help with determining primality of $N$.



                    Famous examples of this include Pépin's Test for primality of numbers of the form $2^{2^n}+1$, and Proth's Theorem for primality of numbers of the form $k times 2^n+1$ where $k<2^n$ (Proth primes feature on the top 20 known primes).



                    There are more general primality tests for $N+1$ based on (partial) knowledge of the factorisation of $N$, but they tend to be less elegant. For example, this was snipped from "Factorizations of $b^n pm 1$, b = 2, 3, 5, 6, 7, 10, 11, 12 Up to High Powers" by Brillhart, Lehmer, Selfridge, Tuckerman, and Wagstaff, Jr.:




                    Theorem 11. Let $N-1=FR$, where $F$ is completely factored and $(F,R)=1$. Suppose there exists an $a$ for which $a^{N-1} equiv 1 pmod N$ and $(a^{(N-1)/q}-1,N)=1$ for each prime factor $q$ of $F$. Let $R=rF+s$, $1 leq s < F$, and suppose $N<2F^3+2F$, $F>2$. If $r$ is odd, or if $r$ is even and $s^2-4r=t^2$, then $N$ is prime. Otherwise, $s^2-4r=t^2$ and $$N = [frac{1}{2}(s-t)F+1][frac{1}{2}(s+t)F+1].$$







                    share|cite|improve this answer
























                      7












                      7








                      7






                      While the factorisation of $N$ might not help much with the factorisation of $N+1$, in special circumstances, it can help with determining primality of $N$.



                      Famous examples of this include Pépin's Test for primality of numbers of the form $2^{2^n}+1$, and Proth's Theorem for primality of numbers of the form $k times 2^n+1$ where $k<2^n$ (Proth primes feature on the top 20 known primes).



                      There are more general primality tests for $N+1$ based on (partial) knowledge of the factorisation of $N$, but they tend to be less elegant. For example, this was snipped from "Factorizations of $b^n pm 1$, b = 2, 3, 5, 6, 7, 10, 11, 12 Up to High Powers" by Brillhart, Lehmer, Selfridge, Tuckerman, and Wagstaff, Jr.:




                      Theorem 11. Let $N-1=FR$, where $F$ is completely factored and $(F,R)=1$. Suppose there exists an $a$ for which $a^{N-1} equiv 1 pmod N$ and $(a^{(N-1)/q}-1,N)=1$ for each prime factor $q$ of $F$. Let $R=rF+s$, $1 leq s < F$, and suppose $N<2F^3+2F$, $F>2$. If $r$ is odd, or if $r$ is even and $s^2-4r=t^2$, then $N$ is prime. Otherwise, $s^2-4r=t^2$ and $$N = [frac{1}{2}(s-t)F+1][frac{1}{2}(s+t)F+1].$$







                      share|cite|improve this answer












                      While the factorisation of $N$ might not help much with the factorisation of $N+1$, in special circumstances, it can help with determining primality of $N$.



                      Famous examples of this include Pépin's Test for primality of numbers of the form $2^{2^n}+1$, and Proth's Theorem for primality of numbers of the form $k times 2^n+1$ where $k<2^n$ (Proth primes feature on the top 20 known primes).



                      There are more general primality tests for $N+1$ based on (partial) knowledge of the factorisation of $N$, but they tend to be less elegant. For example, this was snipped from "Factorizations of $b^n pm 1$, b = 2, 3, 5, 6, 7, 10, 11, 12 Up to High Powers" by Brillhart, Lehmer, Selfridge, Tuckerman, and Wagstaff, Jr.:




                      Theorem 11. Let $N-1=FR$, where $F$ is completely factored and $(F,R)=1$. Suppose there exists an $a$ for which $a^{N-1} equiv 1 pmod N$ and $(a^{(N-1)/q}-1,N)=1$ for each prime factor $q$ of $F$. Let $R=rF+s$, $1 leq s < F$, and suppose $N<2F^3+2F$, $F>2$. If $r$ is odd, or if $r$ is even and $s^2-4r=t^2$, then $N$ is prime. Otherwise, $s^2-4r=t^2$ and $$N = [frac{1}{2}(s-t)F+1][frac{1}{2}(s+t)F+1].$$








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jul 20 '12 at 12:22









                      Douglas S. StonesDouglas S. Stones

                      17.6k454109




                      17.6k454109























                          6














                          As I wrote when this question was raised at MathOverflow, if knowing the factorization of $n$ told you much about the factorization of $n+1$, the Fermat numbers $2^{2^n}+1$ would be easy --- but, they aren't.






                          share|cite|improve this answer




























                            6














                            As I wrote when this question was raised at MathOverflow, if knowing the factorization of $n$ told you much about the factorization of $n+1$, the Fermat numbers $2^{2^n}+1$ would be easy --- but, they aren't.






                            share|cite|improve this answer


























                              6












                              6








                              6






                              As I wrote when this question was raised at MathOverflow, if knowing the factorization of $n$ told you much about the factorization of $n+1$, the Fermat numbers $2^{2^n}+1$ would be easy --- but, they aren't.






                              share|cite|improve this answer














                              As I wrote when this question was raised at MathOverflow, if knowing the factorization of $n$ told you much about the factorization of $n+1$, the Fermat numbers $2^{2^n}+1$ would be easy --- but, they aren't.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Apr 13 '17 at 12:58









                              Community

                              1




                              1










                              answered Jul 20 '12 at 10:38









                              Gerry MyersonGerry Myerson

                              146k8147298




                              146k8147298























                                  4














                                  Finding the connection of the prime factorization of n and n+1 would go hand in hand with proving Collatz Conjecture.






                                  share|cite|improve this answer

















                                  • 3




                                    Would it really?
                                    – Lord Shark the Unknown
                                    May 29 '17 at 20:52










                                  • I already heard that. Can you give details?
                                    – DaBler
                                    Nov 2 '18 at 19:35










                                  • DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                    – Collag3n
                                    Dec 5 '18 at 19:01
















                                  4














                                  Finding the connection of the prime factorization of n and n+1 would go hand in hand with proving Collatz Conjecture.






                                  share|cite|improve this answer

















                                  • 3




                                    Would it really?
                                    – Lord Shark the Unknown
                                    May 29 '17 at 20:52










                                  • I already heard that. Can you give details?
                                    – DaBler
                                    Nov 2 '18 at 19:35










                                  • DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                    – Collag3n
                                    Dec 5 '18 at 19:01














                                  4












                                  4








                                  4






                                  Finding the connection of the prime factorization of n and n+1 would go hand in hand with proving Collatz Conjecture.






                                  share|cite|improve this answer












                                  Finding the connection of the prime factorization of n and n+1 would go hand in hand with proving Collatz Conjecture.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered May 29 '17 at 20:45









                                  NoahNoah

                                  411




                                  411








                                  • 3




                                    Would it really?
                                    – Lord Shark the Unknown
                                    May 29 '17 at 20:52










                                  • I already heard that. Can you give details?
                                    – DaBler
                                    Nov 2 '18 at 19:35










                                  • DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                    – Collag3n
                                    Dec 5 '18 at 19:01














                                  • 3




                                    Would it really?
                                    – Lord Shark the Unknown
                                    May 29 '17 at 20:52










                                  • I already heard that. Can you give details?
                                    – DaBler
                                    Nov 2 '18 at 19:35










                                  • DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                    – Collag3n
                                    Dec 5 '18 at 19:01








                                  3




                                  3




                                  Would it really?
                                  – Lord Shark the Unknown
                                  May 29 '17 at 20:52




                                  Would it really?
                                  – Lord Shark the Unknown
                                  May 29 '17 at 20:52












                                  I already heard that. Can you give details?
                                  – DaBler
                                  Nov 2 '18 at 19:35




                                  I already heard that. Can you give details?
                                  – DaBler
                                  Nov 2 '18 at 19:35












                                  DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                  – Collag3n
                                  Dec 5 '18 at 19:01




                                  DaBler. dividing by 2 or multiplying by 3 only plays with two of the potential prime factors. adding 1 is what makes it interesting
                                  – Collag3n
                                  Dec 5 '18 at 19:01











                                  1














                                  I found a relation, here a proff:



                                  We know that:



                                  $n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



                                  $alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



                                  Then:



                                  $n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



                                  Where:



                                  $beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



                                  In other words:



                                  $n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



                                  $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



                                  Finally, this is the relation:



                                  $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



                                  In summary, Eq.1 can be used as a method for decomposition in prime factors, let's see an example:



                                  $n=60$



                                  $beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



                                  Then:



                                  $beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



                                  $beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



                                  $beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



                                  Finally:



                                  $60=2^{2}3^{1}5^{1}$






                                  share|cite|improve this answer




























                                    1














                                    I found a relation, here a proff:



                                    We know that:



                                    $n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



                                    $alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



                                    Then:



                                    $n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



                                    Where:



                                    $beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



                                    In other words:



                                    $n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



                                    $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



                                    Finally, this is the relation:



                                    $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



                                    In summary, Eq.1 can be used as a method for decomposition in prime factors, let's see an example:



                                    $n=60$



                                    $beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



                                    Then:



                                    $beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



                                    $beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



                                    $beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



                                    Finally:



                                    $60=2^{2}3^{1}5^{1}$






                                    share|cite|improve this answer


























                                      1












                                      1








                                      1






                                      I found a relation, here a proff:



                                      We know that:



                                      $n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



                                      $alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



                                      Then:



                                      $n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



                                      Where:



                                      $beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



                                      In other words:



                                      $n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



                                      $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



                                      Finally, this is the relation:



                                      $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



                                      In summary, Eq.1 can be used as a method for decomposition in prime factors, let's see an example:



                                      $n=60$



                                      $beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



                                      Then:



                                      $beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



                                      $beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



                                      $beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



                                      Finally:



                                      $60=2^{2}3^{1}5^{1}$






                                      share|cite|improve this answer














                                      I found a relation, here a proff:



                                      We know that:



                                      $n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



                                      $alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



                                      Then:



                                      $n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



                                      Where:



                                      $beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



                                      In other words:



                                      $n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



                                      $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



                                      Finally, this is the relation:



                                      $beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



                                      In summary, Eq.1 can be used as a method for decomposition in prime factors, let's see an example:



                                      $n=60$



                                      $beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



                                      Then:



                                      $beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



                                      $beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



                                      $beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



                                      Finally:



                                      $60=2^{2}3^{1}5^{1}$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 5 '18 at 3:40

























                                      answered Dec 4 '18 at 19:48









                                      Mauricio AreizaMauricio Areiza

                                      143




                                      143






























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