Prove that it does not exist two number such that $m^2+n^2=6 underbrace {0 cdots 0}$











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Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?






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    Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?






    modular-arithmetic divisibility







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      Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?






      modular-arithmetic divisibility







      share|cite|improve this question















      Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?






      modular-arithmetic divisibility




      modular-arithmetic divisibility






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      edited Nov 13 at 5:00

























      asked Nov 13 at 4:51









      Marko Škorić

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          $60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.






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            Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
            Prime factorizaton of $6underbrace {0cdots 0}$ is



            $$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$



            Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers






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              $60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.






              share|cite|improve this answer

























                up vote
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                $60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.






                share|cite|improve this answer























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                  up vote
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                  $60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.






                  share|cite|improve this answer












                  $60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.







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                  answered Nov 13 at 4:57









                  Robert Israel

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                      Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
                      Prime factorizaton of $6underbrace {0cdots 0}$ is



                      $$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$



                      Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers






                      share|cite|improve this answer










                      New contributor




                      Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        up vote
                        0
                        down vote













                        Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
                        Prime factorizaton of $6underbrace {0cdots 0}$ is



                        $$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$



                        Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers






                        share|cite|improve this answer










                        New contributor




                        Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          up vote
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                          up vote
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                          down vote









                          Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
                          Prime factorizaton of $6underbrace {0cdots 0}$ is



                          $$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$



                          Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers






                          share|cite|improve this answer










                          New contributor




                          Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
                          Prime factorizaton of $6underbrace {0cdots 0}$ is



                          $$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$



                          Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers







                          share|cite|improve this answer










                          New contributor




                          Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 14 hours ago





















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                          answered 17 hours ago









                          Samurai

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