China Western Mathematical Olympiad

$$x^2+y^2\equiv 0,1,2(\mathrm{mod}4).$$ Let $n\in S$. By the above equality, we get $n\equiv 1(\mathrm{mod}4)$. Thus, we may assume that $$n-1=a^2+b^2,a\ge b,$$ $$n=c^2+d^2,c>d,$$ $$n+1=e^2+f^2,e\ge f,$$ where $a,b,c,d,e,f$ are positive integers. Therefore $$n^2+1=n^2+1^2,$$ $$n^2=(c^2+d^2{)}^2=(c^2-d^2{)}^2+(2cd{)}^2,$$ $$n^2-1=(a^2+b^2)(e^2+f^2)=(ae-bf{)}^2+(af+be{)}^2.$$ Suppose that $b=a$ and $f=e$, we have $n-1=2a^2$, $n+1=2e^2$. Taking the difference of these two equations yields $e^2-a^2=1$. Then, $e-a\ge 1$. But $$1=e^2-a^2=(e+a)(e-a)>1,$$ a contradiction! So $b=a$ and $f=e$ do not happen simultaneously. Therefore $ae-bf>0$, and $n^2\in S$.