67th Romanian Mathematical Olympiad

a) Suppose, by way of contradiction, that one can find rational numbers $x$, $y$, $z$ such that $7=x^2+y^2+z^2$. Writing $x$, $y$, $z$ as fractions and clearing denominators yield an equality like $$7n^2=a^2+b^2+c^2,(\ast )$$ where $n$, $a$, $b$, $c$ are nonnegative integers, not all zero. If $n$ is even, then $a$, $b$, $c$ are all even, as well (obviously, they cannot be all odd, and if exactly two of them are odd, the sum of their squares equals $2(\mathrm{mod}4)$, while $7n^2$ equals $0(\mathrm{mod}4)$). Dividing by $4$ yields $$7n_1^2=a_1^2+b_1^2+c_1^2,$$ and, clearly, $0<a_1+b_1+c_1<a+b+c$. If $n_1$ is still even, we repeat the previous transformation (this can only happen finitely many times). If $n$ is odd, say $n=2k+1$, then $7n^2=7\cdot 4k(k+1)+7=M8+7$. Since the remainder of a square when divided by $8$ equals $0$, $1$ or $4$, we deduce that equality $(\ast )$ is impossible.

b) We induct on $m$. The base case $m=1$ follows from the hypothesis. We assume the statement true for all $m\le n$ and prove it for $n+1$. If $n+1=2k$, then $k\le n$, hence $a^k$ can be written as a sum of squares of three rational numbers, for instance $a^k=x^2+y^2+z^2$, where $x\ge y\ge z$. Then $$a^{2k}=(x^2+y^2+z^2{)}^2=(x^2+y^2-z^2{)}^2+(2xz{)}^2+(2yz{)}^2,$$ and the numbers $x^2+y^2-z^2$, $2xz$, $2yz$ are obviously rational numbers.

If $n+1=2k+1$, using $a=x^2+y^2+z^2$, we get $$a^{n+1}=a^{2k}\cdot a=a^{2k}(x^2+y^2+z^2)=(a^{k}x{)}^2+(a^{k}y{)}^2+(a^{k}z{)}^2,$$