IMAR Mathematical Competition

() Let $(p,q,r)$ be a Pythagorean triple of positive integers, $p^2+q^2=r^2$, such that $p\equiv -1(\mathrm{mod}4)$, $q\equiv 0(\mathrm{mod}4)$, $r\equiv 1(\mathrm{mod}4)$, and $p<q$. Then every positive integer $N\equiv 3(\mathrm{mod}8)$, that is divisible by $r^2$, is the sum of three odd squares whose positive square roots are not congruent modulo $4$. --- Consequently, a positive integer $n\equiv \frac{3(r^2-1)}{8}(\mathrm{mod}{r}^2)$ 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 $N/r^2\equiv 3(\mathrm{mod}8)$, 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 $(ru-rv)+((pu+qv)-(qu-pv))\equiv (u-v)+(u+v)\equiv 2u\equiv 2(\mathrm{mod}4)$, 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.