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
$$7n^2 = a^2 + b^2 + c^2, \quad (*)$$
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 $2 \pmod{4}$, while $7n^2$ equals $0 \pmod{4}$). 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 $(*)$ 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
$$a^{n+1} = a^{2k} \cdot a = a^{2k} (x^2 + y^2 + z^2) = (a^k x)^2 + (a^k y)^2 + (a^k z)^2,$$