IMAR Mathematical Competition

(*) Let $(p, q, r)$ be a Pythagorean triple of positive integers, $p^2 + q^2 = r^2$, such that $p \equiv -1 \pmod{4}$, $q \equiv 0 \pmod{4}$, $r \equiv 1 \pmod{4}$, and $p < q$. Then every positive integer $N \equiv 3 \pmod{8}$, that is divisible by $r^2$, is the sum of three odd squares whose positive square roots are not congruent modulo $4$.
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Consequently, a positive integer $n \equiv \frac{3(r^2 - 1)}{8} \pmod{r^2}$ is expressible in the form $n = \frac{a(a+1)}{2} + \frac{b(b+1)}{2} + \frac{c(c+1)}{2}$ for some non-negative integers $a, b, c$ that do not share parity.
The problem at hand is the special case where $(p, q, r) = (3, 4, 5)$.
To prove (*), notice that $N/r^2 \equiv 3 \pmod{8}$, so it is not of the form $4^k(8\ell + 7)$, and is therefore a sum of three odd squares (Gauss-Legendre).
Write $N = (ru)^2 + (rv)^2 + (rw)^2$ for some positive odd integers $u, v, w$, and assume, without loss of generality, that $u \ge v$, to write $(ru)^2 + (rv)^2 = (pu + qv)^2 + (qu - pv)^2$.
Since $(ru - rv) + ((pu + qv) - (qu - pv)) \equiv (u - v) + (u + v) \equiv 2u \equiv 2 \pmod{4}$, the entries of one of the pairs of positive odd integers $(ru, rv), (pu + qv, qu - pv)$ are not congruent modulo $4$. This ends the proof.