Can 2018 be written as a sum of two squares? If can, what is the expression?
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Can $2018$ be written as a sum of two squares?
If it can be, what are the numbers?
I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.
elementary-number-theory
edited Dec 16 '17 at 17:39
amWhy
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
add a comment |
up vote
0
down vote
favorite
Can $2018$ be written as a sum of two squares?
If it can be, what are the numbers?
I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.
elementary-number-theory
edited Dec 16 '17 at 17:39
amWhy
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
1
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can $2018$ be written as a sum of two squares?
If it can be, what are the numbers?
I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.
elementary-number-theory
edited Dec 16 '17 at 17:39
amWhy
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
Can $2018$ be written as a sum of two squares?
If it can be, what are the numbers?
I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.
elementary-number-theory
elementary-number-theory
edited Dec 16 '17 at 17:39
amWhy
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
edited Dec 16 '17 at 17:39
amWhy
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
edited Dec 16 '17 at 17:39
amWhy
191k27223439
edited Dec 16 '17 at 17:39
amWhy
191k27223439
edited Dec 16 '17 at 17:39
amWhy
191k27223439
191k27223439
asked Dec 7 '17 at 4:15
Joe sae
65
asked Dec 7 '17 at 4:15
Joe sae
65
asked Dec 7 '17 at 4:15
Joe sae
65
65
1
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41
add a comment |
1
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41
1
1
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41
add a comment |
1 Answer
1
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Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.
The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$
A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.
edited Nov 25 at 19:49
LoPablo
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.
The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$
A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.
edited Nov 25 at 19:49
LoPablo
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
add a comment |
up vote
5
down vote
Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.
The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$
A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.
edited Nov 25 at 19:49
LoPablo
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
add a comment |
up vote
5
down vote
up vote
5
down vote
Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.
The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$
A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.
edited Nov 25 at 19:49
LoPablo
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.
The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$
A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.
edited Nov 25 at 19:49
LoPablo
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
edited Nov 25 at 19:49
LoPablo
33
edited Nov 25 at 19:49
LoPablo
33
edited Nov 25 at 19:49
LoPablo
33
33
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
answered Dec 7 '17 at 4:51
achille hui
94.8k5129255
94.8k5129255
add a comment |
add a comment |
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1
If all you care about is the numbers, then wolframalpha can help.
– JMoravitz
Dec 7 '17 at 4:18
By hit and trial, you can verify that $2018 = 13^2 + cdots$
– Math Lover
Dec 7 '17 at 4:21
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
– Ross Millikan
Dec 7 '17 at 4:21
See math.stackexchange.com/questions/1828694/…
– lab bhattacharjee
Dec 7 '17 at 4:41