Proof that $mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})$ is not a subfield of $mathbb{R}$












1














Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know.



$alpha:=sqrt{2}+sqrt{3}notin U:= mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})iffalpha notin mathbb{Q}(sqrt{2})$ and $alpha notin mathbb{Q}(sqrt{3})$



Proof: $alpha notin mathbb{Q}(sqrt{2})$



Assume $exists_{r,sin mathbb{Q}}r+ssqrt{2}=sqrt{2}+sqrt{3}$



$iff sqrt{3}=r+(s-1)sqrt2$



$iff 3=r^2+2r(s-1)sqrt2+2(s-1)^2$



Which implicates $3=0$ or $sqrt3in mathbb{Q}$ or $sqrt{3/2}in mathbb{Q}$ or $sqrt{2}in mathbb{Q}$



I don't understand the last implication, I Need some help here.



Many Thanks






abstract-algebra







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  • I would really like to understand the last implication
    – RM777
    Nov 30 at 9:51
















1














Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know.



$alpha:=sqrt{2}+sqrt{3}notin U:= mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})iffalpha notin mathbb{Q}(sqrt{2})$ and $alpha notin mathbb{Q}(sqrt{3})$



Proof: $alpha notin mathbb{Q}(sqrt{2})$



Assume $exists_{r,sin mathbb{Q}}r+ssqrt{2}=sqrt{2}+sqrt{3}$



$iff sqrt{3}=r+(s-1)sqrt2$



$iff 3=r^2+2r(s-1)sqrt2+2(s-1)^2$



Which implicates $3=0$ or $sqrt3in mathbb{Q}$ or $sqrt{3/2}in mathbb{Q}$ or $sqrt{2}in mathbb{Q}$



I don't understand the last implication, I Need some help here.



Many Thanks






abstract-algebra







share|cite|improve this question






















  • I would really like to understand the last implication
    – RM777
    Nov 30 at 9:51














1












1








1







Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know.



$alpha:=sqrt{2}+sqrt{3}notin U:= mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})iffalpha notin mathbb{Q}(sqrt{2})$ and $alpha notin mathbb{Q}(sqrt{3})$



Proof: $alpha notin mathbb{Q}(sqrt{2})$



Assume $exists_{r,sin mathbb{Q}}r+ssqrt{2}=sqrt{2}+sqrt{3}$



$iff sqrt{3}=r+(s-1)sqrt2$



$iff 3=r^2+2r(s-1)sqrt2+2(s-1)^2$



Which implicates $3=0$ or $sqrt3in mathbb{Q}$ or $sqrt{3/2}in mathbb{Q}$ or $sqrt{2}in mathbb{Q}$



I don't understand the last implication, I Need some help here.



Many Thanks






abstract-algebra







share|cite|improve this question













Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know.



$alpha:=sqrt{2}+sqrt{3}notin U:= mathbb{Q}(sqrt{2})cupmathbb{Q}(sqrt{3})iffalpha notin mathbb{Q}(sqrt{2})$ and $alpha notin mathbb{Q}(sqrt{3})$



Proof: $alpha notin mathbb{Q}(sqrt{2})$



Assume $exists_{r,sin mathbb{Q}}r+ssqrt{2}=sqrt{2}+sqrt{3}$



$iff sqrt{3}=r+(s-1)sqrt2$



$iff 3=r^2+2r(s-1)sqrt2+2(s-1)^2$



Which implicates $3=0$ or $sqrt3in mathbb{Q}$ or $sqrt{3/2}in mathbb{Q}$ or $sqrt{2}in mathbb{Q}$



I don't understand the last implication, I Need some help here.



Many Thanks






abstract-algebra




abstract-algebra






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asked Nov 30 at 9:34









RM777

1908




1908












  • I would really like to understand the last implication
    – RM777
    Nov 30 at 9:51


















  • I would really like to understand the last implication
    – RM777
    Nov 30 at 9:51
















I would really like to understand the last implication
– RM777
Nov 30 at 9:51




I would really like to understand the last implication
– RM777
Nov 30 at 9:51










4 Answers
4






active

oldest

votes


















2














We know that any element of $mathbb{Q}[sqrt{2}]$ can be uniquely written as $a + b sqrt{2}$ with $a, b in mathbb{Q}$ (the unicity is important). So, if $alpha in mathbb{Q}[sqrt{2}]$, then so does $alpha - sqrt{2} = sqrt{3} in mathbb{Q}[sqrt{2}]$. So you can write $sqrt{3} = r + s sqrt{2}$. Squaring that equality, we get



$$3 = (r^2 + 2 s^2) + 2rs sqrt{2} $$



By unicity, the number $3$ can be uniquely written as $3 + 0 sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $sqrt{3}$ is not in $mathbb{Q}[sqrt{2}]$.






share|cite|improve this answer





























    3














    Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
    On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.






    share|cite|improve this answer





















    • This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
      – Joel Cohen
      Nov 30 at 10:03










    • Indeed, but the cases either $a=0$ or $b=0$ are easy.
      – Wuestenfux
      Nov 30 at 10:05



















    3














    Me neither - but if $r(s-1) neq 0$ you can rewrite the last equation to obtain $$mathbb{Q} ni frac{3-r^2-2(s-1)^2}{2r(s-1)} = sqrt{2} notin mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.






    share|cite|improve this answer























    • You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
      – Joel Cohen
      Nov 30 at 10:02










    • You're right of course, edited my answer.
      – Stockfish
      Nov 30 at 13:25



















    2














    Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?



    Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.



    So definitely it cannot be a Subfield of $mathbb{R}$.






    share|cite|improve this answer





















    • Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
      – RM777
      Nov 30 at 9:43










    • Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
      – Sujit Bhattacharyya
      Nov 30 at 9:46













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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    We know that any element of $mathbb{Q}[sqrt{2}]$ can be uniquely written as $a + b sqrt{2}$ with $a, b in mathbb{Q}$ (the unicity is important). So, if $alpha in mathbb{Q}[sqrt{2}]$, then so does $alpha - sqrt{2} = sqrt{3} in mathbb{Q}[sqrt{2}]$. So you can write $sqrt{3} = r + s sqrt{2}$. Squaring that equality, we get



    $$3 = (r^2 + 2 s^2) + 2rs sqrt{2} $$



    By unicity, the number $3$ can be uniquely written as $3 + 0 sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $sqrt{3}$ is not in $mathbb{Q}[sqrt{2}]$.






    share|cite|improve this answer


























      2














      We know that any element of $mathbb{Q}[sqrt{2}]$ can be uniquely written as $a + b sqrt{2}$ with $a, b in mathbb{Q}$ (the unicity is important). So, if $alpha in mathbb{Q}[sqrt{2}]$, then so does $alpha - sqrt{2} = sqrt{3} in mathbb{Q}[sqrt{2}]$. So you can write $sqrt{3} = r + s sqrt{2}$. Squaring that equality, we get



      $$3 = (r^2 + 2 s^2) + 2rs sqrt{2} $$



      By unicity, the number $3$ can be uniquely written as $3 + 0 sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $sqrt{3}$ is not in $mathbb{Q}[sqrt{2}]$.






      share|cite|improve this answer
























        2












        2








        2






        We know that any element of $mathbb{Q}[sqrt{2}]$ can be uniquely written as $a + b sqrt{2}$ with $a, b in mathbb{Q}$ (the unicity is important). So, if $alpha in mathbb{Q}[sqrt{2}]$, then so does $alpha - sqrt{2} = sqrt{3} in mathbb{Q}[sqrt{2}]$. So you can write $sqrt{3} = r + s sqrt{2}$. Squaring that equality, we get



        $$3 = (r^2 + 2 s^2) + 2rs sqrt{2} $$



        By unicity, the number $3$ can be uniquely written as $3 + 0 sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $sqrt{3}$ is not in $mathbb{Q}[sqrt{2}]$.






        share|cite|improve this answer












        We know that any element of $mathbb{Q}[sqrt{2}]$ can be uniquely written as $a + b sqrt{2}$ with $a, b in mathbb{Q}$ (the unicity is important). So, if $alpha in mathbb{Q}[sqrt{2}]$, then so does $alpha - sqrt{2} = sqrt{3} in mathbb{Q}[sqrt{2}]$. So you can write $sqrt{3} = r + s sqrt{2}$. Squaring that equality, we get



        $$3 = (r^2 + 2 s^2) + 2rs sqrt{2} $$



        By unicity, the number $3$ can be uniquely written as $3 + 0 sqrt{2}$, so in the right hand side we get $r^2 + 2 s^2 = 3$ and $2rs=0$. From $rs = 0$ we get that $r=0$ or $s = 0$. If $s=0$, then $r^2 = 3$, which is impossible since $r$ is rational. And if $r = 0$, then $2s^2=3$ so $s^2 = frac{3}{2}$, which is also impossible since $s$ is rational. So in summary, $sqrt{3}$ is not in $mathbb{Q}[sqrt{2}]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 at 9:54









        Joel Cohen

        7,28412037




        7,28412037























            3














            Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
            On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.






            share|cite|improve this answer





















            • This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
              – Joel Cohen
              Nov 30 at 10:03










            • Indeed, but the cases either $a=0$ or $b=0$ are easy.
              – Wuestenfux
              Nov 30 at 10:05
















            3














            Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
            On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.






            share|cite|improve this answer





















            • This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
              – Joel Cohen
              Nov 30 at 10:03










            • Indeed, but the cases either $a=0$ or $b=0$ are easy.
              – Wuestenfux
              Nov 30 at 10:05














            3












            3








            3






            Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
            On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.






            share|cite|improve this answer












            Well, $alphanotin {Bbb Q}(sqrt 2)$ is equivalent to $sqrt 3notin {Bbb Q}(sqrt 2)$.
            On the contrary, $sqrt 3 = a+bsqrt 2$ for some rational numbers $a,b$. Squaring gives $3 = a^2+2absqrt 2 + 2b^2$. Then $sqrt 2$ could be written as a rational number, which is impossible as it is irrational.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 30 at 9:43









            Wuestenfux

            3,2211410




            3,2211410












            • This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
              – Joel Cohen
              Nov 30 at 10:03










            • Indeed, but the cases either $a=0$ or $b=0$ are easy.
              – Wuestenfux
              Nov 30 at 10:05


















            • This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
              – Joel Cohen
              Nov 30 at 10:03










            • Indeed, but the cases either $a=0$ or $b=0$ are easy.
              – Wuestenfux
              Nov 30 at 10:05
















            This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
            – Joel Cohen
            Nov 30 at 10:03




            This only holds if you assume $ab ne 0$. You still have to consider the case $ab=0$ and derive a contradiction from that.
            – Joel Cohen
            Nov 30 at 10:03












            Indeed, but the cases either $a=0$ or $b=0$ are easy.
            – Wuestenfux
            Nov 30 at 10:05




            Indeed, but the cases either $a=0$ or $b=0$ are easy.
            – Wuestenfux
            Nov 30 at 10:05











            3














            Me neither - but if $r(s-1) neq 0$ you can rewrite the last equation to obtain $$mathbb{Q} ni frac{3-r^2-2(s-1)^2}{2r(s-1)} = sqrt{2} notin mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.






            share|cite|improve this answer























            • You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
              – Joel Cohen
              Nov 30 at 10:02










            • You're right of course, edited my answer.
              – Stockfish
              Nov 30 at 13:25
















            3














            Me neither - but if $r(s-1) neq 0$ you can rewrite the last equation to obtain $$mathbb{Q} ni frac{3-r^2-2(s-1)^2}{2r(s-1)} = sqrt{2} notin mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.






            share|cite|improve this answer























            • You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
              – Joel Cohen
              Nov 30 at 10:02










            • You're right of course, edited my answer.
              – Stockfish
              Nov 30 at 13:25














            3












            3








            3






            Me neither - but if $r(s-1) neq 0$ you can rewrite the last equation to obtain $$mathbb{Q} ni frac{3-r^2-2(s-1)^2}{2r(s-1)} = sqrt{2} notin mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.






            share|cite|improve this answer














            Me neither - but if $r(s-1) neq 0$ you can rewrite the last equation to obtain $$mathbb{Q} ni frac{3-r^2-2(s-1)^2}{2r(s-1)} = sqrt{2} notin mathbb{Q},$$ a contradiction. If $s=1$ we have $3=r^2$, if $r=0$ the equation reads $3=2(s-1)^2$ - both is impossible because of the uniqueness of prime factorization.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 30 at 13:29

























            answered Nov 30 at 9:43









            Stockfish

            51726




            51726












            • You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
              – Joel Cohen
              Nov 30 at 10:02










            • You're right of course, edited my answer.
              – Stockfish
              Nov 30 at 13:25


















            • You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
              – Joel Cohen
              Nov 30 at 10:02










            • You're right of course, edited my answer.
              – Stockfish
              Nov 30 at 13:25
















            You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
            – Joel Cohen
            Nov 30 at 10:02




            You have to assume that $r(s-1) ne 0$ to do that. Which then leaves the case $r(s-1) = 0$ to be considered.
            – Joel Cohen
            Nov 30 at 10:02












            You're right of course, edited my answer.
            – Stockfish
            Nov 30 at 13:25




            You're right of course, edited my answer.
            – Stockfish
            Nov 30 at 13:25











            2














            Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?



            Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.



            So definitely it cannot be a Subfield of $mathbb{R}$.






            share|cite|improve this answer





















            • Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
              – RM777
              Nov 30 at 9:43










            • Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
              – Sujit Bhattacharyya
              Nov 30 at 9:46


















            2














            Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?



            Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.



            So definitely it cannot be a Subfield of $mathbb{R}$.






            share|cite|improve this answer





















            • Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
              – RM777
              Nov 30 at 9:43










            • Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
              – Sujit Bhattacharyya
              Nov 30 at 9:46
















            2












            2








            2






            Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?



            Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.



            So definitely it cannot be a Subfield of $mathbb{R}$.






            share|cite|improve this answer












            Instead of that just try to check is this $U:=mathbb{Q}(sqrt 2)cup mathbb{Q}(sqrt 3)$ a Group?



            Let us take $sqrt2,sqrt3in U$ [Both exists] but $sqrt2+sqrt3notin U$.



            So definitely it cannot be a Subfield of $mathbb{R}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 30 at 9:41









            Sujit Bhattacharyya

            945317




            945317












            • Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
              – RM777
              Nov 30 at 9:43










            • Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
              – Sujit Bhattacharyya
              Nov 30 at 9:46




















            • Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
              – RM777
              Nov 30 at 9:43










            • Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
              – Sujit Bhattacharyya
              Nov 30 at 9:46


















            Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
            – RM777
            Nov 30 at 9:43




            Yes that's exactly what I want to prove $sqrt2+sqrt3notin U $
            – RM777
            Nov 30 at 9:43












            Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
            – Sujit Bhattacharyya
            Nov 30 at 9:46






            Alright! Note that: $sqrt3notinmathbb{Q}(sqrt2)={a+bsqrt2:a,binmathbb{Q}}$ and so $sqrt2+sqrt3notinmathbb{Q}(sqrt2)$ Further, if we assume $sqrt3=a+bsqrt2$ the $3=a^2+2b^2$ and $2ab=0$ gives the contradiction.
            – Sujit Bhattacharyya
            Nov 30 at 9:46




















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