Let $a,b \in \mathbb{N}$ such that when $a^2+b^2$ is divided by $a+b$, the quotient is $q$ and remainder is $r$ such that $q^2+r=2000$. Find $q,r$
My try: Obviously $a \ne b$ for if we have $r=0$ $\implies$ $q=\sqrt{2000} \notin \mathbb{N}$
Without loss of generality, let $a>b$ and $a-b=c$
Then we have $$a^2+b^2=(a-b)(a+b)+2b^2$$
So $q=a-b, r=2b^2$
Using $q^2+r=2000$ $$c^2+2b^2=2000$$ But i guess we cannot find two natural numbers $c,b$ satisfying above equation. Any help?
From $$a^2+b^2=q(a+b)+r\ ,\qquad q^2+r=2000$$ we get $$q^2-(a+b)q+(a^2+b^2-2000)=0\ .$$ For this to have real solutions we need $$(a+b)^2\ge 4(a^2+b^2-2000)$$ which simplifies to $$8000\ge 3a^2-2ab+3b^2=2a^2+2b^2+(a-b)^2\ .$$ Therefore $8000>2a^2,\,2b^2$ and so $$a,b\le63\ .$$ However the largest square below $2000$ is $44^2$. If $q\ne44$ then $$r=2000-q^2\ge2000-43^2=151\ .$$ This is impossible if we require $r<a+b$. So the only possibility is $q=44$, $r=64$ and $$a^2+b^2=44(a+b)+64\ .$$ Completing the square, $$(a-22)^2+(b-22)^2=1032\ .$$ Now $3$ is a factor of $1032$, and $3\equiv3\pmod4$, and $3^2\not\mid1032$, so $1032$ is not a sum of two squares and there is no solution.