orl wrote: 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.$ We note $ r < q$ Finding possible values of $ q$. Assume $ r = 0$. Hence $ q_0 = \sqrt{1977}>\sqrt{1936}=44$ Therefore, the only value of $ q$ must be $ \leq 44$, since $ 1977-43^2>43$, $ q=44$ Therefore, $ r=41$ Therefore, we solve $ 44(a+b)+41=a^2+b^2$ to find $ (a,b)=(50, 7), (7,50), (50, 37), (37, 50)$ (I had to case bass that last bit, anyone got a better proof?)

why is $ r < q$ i was under the impression that the task states $ a^2 + b^2 = r\mod(a + b)$ not $ a^2 + b^2 = r\mod q$ and therefore only $ r < a + b$ not $ r < q$ you however got the right result obivouslly $ q<45$ $ q^2-q(a+b)+a^2+b^2-1977=0$ therefore $ 7908-3a^2-3b^2+2ab$ must be a perfect square. so $ 3a^2+3b^2-2ab=3(a+b)^2 -8ab\leq7908$ By AM-GM we get $ 3(a+b)^2-8ab\geq3(a+b)^2-8(\frac{a+b}{2})^2=(a+b)^2>7908$ except if $ a+b\leq88$ and since $ r<a+b<89$ and $ 1977-43^2=128$ we can conclude $ q=44$ $ r=41$

Jure the frEEEk wrote: why is $ r < q$ i was under the impression that the task states $ a^2 + b^2 = r\mod(a + b)$ not $ a^2 + b^2 = r\mod q$ and therefore only $ r < a + b$ not $ r < q$ Yep, thats a screw up on my part

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. $\blacksquare$ Arkan

Note $r > 0$ as 1977 is not a square number By QM-AM, we have $a^2 + b^2 \ge \frac{(a + b)^2}{2}$ $q(a + b) + r \ge \frac{(a + b)^2}{2}$ $2q(a + b) > (a + b)^2$ $2q > a + b > r$ $(q + 1)^2 > a + b + q^2 > q^2 + r = 1977$ so $q = 44$, then $r = 41$