15. (IRE 1) Let $a, b, c, d, m, n$ be positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}=$ 1989, $a+b+c+d=m^{2}$, and the largest of $a, b, c, d$ is $n^{2}$. Determine, with proof, the values of $m$ and $n$.
15. By Cauchy's inequality, $44a^{2}+b^{2}+c^{2}=$ $1989-d^{2}$, and so $d^{2}-49 d+206>0$. This inequality does not hold for $5 \leq d \leq 44$. Since $d \geq \sqrt{1989 / 4}>22, d$ must be at least 45 , which is impossible because $45^{2}>1989$. Thus we must have $m^{2}=81$ and $m=9$. Now, $4 d>81$ implies $d \geq 21$. On the other hand, $d361=(p+q+r)^{2}$, a contradiction. Therefore $d=36$ and $n=6$.
Remark. A little more calculation yields the unique solution $a=12$, $b=15, c=18, d=36$.