WakeUp wrote: Let $a$ and $b$ be rational numbers such that $s=a+b=a^2+b^2$. Prove that $s$ can be written as a fraction where the denominator is relatively prime to $6$. It's easy to see that the required property is true if $ab=0$ (since either $s=0$, either $s=1$). So let $a=\frac uv$ and $b=\frac wt$, with $u,v,w,t\ne 0$. we get : $s=\frac uv+\frac wt$ and so $svt=ut+vw$ and so $s^2v^2t^2=(m+n)^2$ where $m=ut$ and $n=vw$ $s=\frac {u^2}{v^2}+\frac {w^2}{t^2}$ and so $sv^2t^2=m^2+n^2$ And so $s=\frac{(m+n)^2}{m^2+n^2}$ If $m^2+n^2$ is even : either both $m,n$ are even and we can replace $m,n$ with $\frac m2,\frac n2$ either both $m,n$ are odd and $m^2+n^2\equiv 2\pmod 4$ while $(m+n)^2\equiv 0\pmod 4$ and we can divide both parts by $2$ and get a fraction with odd denominator. If $m^2+n^2\equiv 0\pmod 3$ : both $m,n$ are divisible by $3$ and we can replace $m,n$ with $\frac m3,\frac n3$ So we always can write $s$ as a fraction whose denominator is odd not divisible by $3$ Q.E.D.