Jumbler wrote: Let $S=\{n|n-1,n,n+1$ can be expressed as the sum of the square of two positive integers.$\}$. Prove that if $n$ in $S$, $n^{2}$ is also in $S$. It is easy. If $n-1=x^{2}+y^{2},n=a^{2}+b^{2},n+1=u^{2}+v^{2}$, then $n^{2}-1=(xu-yv)^{2}+(xv+yu)^{2}, n^{2}=n^{2}+0^{2}=(a^{2}-b^{2})^{2}+(2ab)^{2},n^{2}+1=(n)^{2}+(1)^{2}.$

Rust wrote: Jumbler wrote: Let $S=\{n|n-1,n,n+1$ can be expressed as the sum of the square of two positive integers.$\}$. Prove that if $n$ in $S$, $n^{2}$ is also in $S$. It is easy. If $n-1=x^{2}+y^{2},n=a^{2}+b^{2},n+1=u^{2}+v^{2}$, then $n^{2}-1=(xu-yv)^{2}+(xv+yu)^{2}, n^{2}=n^{2}+0^{2}=(a^{2}-b^{2})^{2}+(2ab)^{2},n^{2}+1=(n)^{2}+(1)^{2}.$ 0 is not positive.

We know that if $n$ is a positive intenger then there exist intengers $x$ and $y$ such that $n=x^{2}+y^{2}$ if each prime factor of $n$ of the form $4k+3$ appears an even number of times. So if $n \in S$ we have that $n-1$,$n$ and $n+1$ satisfies the propertie above.So if each prime factor of $n-1$ of the form $4k+3$ appears an even number of times, and each prime factor of $n+1$ of the form $4k+3$ appears an even number of times we have that each prime factor of $(n+1)(n-1)=n^{2}-1$ of the form $4k+3$ appears an even number of times,thus $n^{2}-1$can be expressed as a sum of two squares.Of course the same is true for $n^{2}$ since each prime factor of it of the form $4k+3$ will appears an even number of times too!.So we just have to prove now that $n^{2}+1$ can be expressed as a sum of two squares.But that is obviously since $n$ is intenger and $1=1^{2}$!!!. So we solved the problem! ........................... Gabriel Ponce

gabriel ponce wrote: We know that if $n$ is a positive intenger then there exist intengers $x$ and $y$ such that $n=x^{2}+y^{2}$ if each prime factor of $n$ of the form $4k+3$ appears an even number of times. So if $n \in S$ we have that $n-1$,$n$ and $n+1$ satisfies the propertie above.So if each prime factor of $n-1$ of the form $4k+3$ appears an even number of times, and each prime factor of $n+1$ of the form $4k+3$ appears an even number of times we have that each prime factor of $(n+1)(n-1)=n^{2}-1$ of the form $4k+3$ appears an even number of times,thus $n^{2}-1$can be expressed as a sum of two squares.Of course the same is true for $n^{2}$ since each prime factor of it of the form $4k+3$ will appears an even number of times too!.So we just have to prove now that $n^{2}+1$ can be expressed as a sum of two squares.But that is obviously since $n$ is intenger and $1=1^{2}$!!!. So we solved the problem! ........................... Gabriel Ponce actualy.....we know that $n$ could be written as a sum of squares if each prime factor of the form "4k+3"appears an even number of times but it is not true that if every prime factor of that form appears an even number of times then the number can really be written like a sum of squares.see for example $7^{2}$. $7=4*1+3$ and it appears an even number of times but $7^{2}$ can not be written that way.but i think rust's solution is the shorteest way to kill the problem which is pretty easy in fact.

ali666 wrote: Rust wrote: Jumbler wrote: Let $S=\{n|n-1,n,n+1$ can be expressed as the sum of the square of two positive integers.$\}$. Prove that if $n$ in $S$, $n^{2}$ is also in $S$. It is easy. If $n-1=x^{2}+y^{2},n=a^{2}+b^{2},n+1=u^{2}+v^{2}$, then $n^{2}-1=(xu-yv)^{2}+(xv+yu)^{2}, n^{2}=n^{2}+0^{2}=(a^{2}-b^{2})^{2}+(2ab)^{2},n^{2}+1=(n)^{2}+(1)^{2}.$ 0 is not positive. $n^{2}=(a^{2}-b^{2})^{2}+(2ab)^{2}$. If $a=b$, then $x^{2}+y^{2}=2a^{2}-1$, $u^{2}+v^{2}=2a^{2}+1$ give contradition, because one of them 3(mod 4).