1977 IMO Problems/Problem 4

Problem

Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$ , we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$

Solution

Using $r=1977-q^2$ , we have $a^2+b^2=(a+b)q+1977-q^2$ , or $q^2-(a+b)q+a^2+b^2-1977=0$ , which implies $\Delta=7908+2ab-2(a^2+b^2)\ge 0$ . If we now assume Wlog that $a\ge b$ , it follows $a+b\le 88$ . If $q\le 43$ , then $r=1977-q^2\ge 128$ , contradicting $r<a+b\le 88$ . But $q\le 44$ from $q^2+r=1977$ , thus $q=44$ . It follows $r=41$ , and we get $a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}$ . By Jacobi's two squares theorem, we infer that $15^2+28^2=1009$ is the only representation of $1009$ as a sum of squares. This forces $\boxed{(a,b)=(37,50) , (7, 50)}$ , and permutations. $\blacksquare$

The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here:

1977 IMO (Problems) • Resources
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1977 IMO Problems/Problem 4

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