Is it a Diophantine?

Let $a, b \in \mathbb{N}$ satisfy the given equation. It is not possible that $a=b$ (since it leads to $a^2+2=2a$ ), so we assume w.l.o.g. that $a>b$ . Next, for $a>b=1$ the equation becomes $a^2 = 2a$ , and one obtains a solution $(a, b) = (2,1)$ .

Let $b>1$ . If $\left[\frac{a^2}{b} \right]=\alpha$ and $\left[\frac{b^2}{a} \right]=\beta$ , then we trivially have $ab \ge \alpha \beta$ . Since also $\frac{a^2+b^2}{ab}\ge2$ , we obtain $\alpha +\beta \ge \alpha \beta +2$ , or equivalently $(\alpha -1)(\beta -1)\le -1$ . But $\alpha \ge 1$ , and therefore $\beta = 0$ . It follows that $a>b^2$ , i.e., $a =b^2+c$ for some $c >0$ . Now the given equation becomes $b^3+2bc+\left[\frac{c^2}{b} \right]=\left[\frac{b^4+2b^2c+b^2+c^2}{b^3+bc} \right]+b^3+bc \\ (c-1)b+\left[\frac{c^2}{b} \right]=\left[\frac{b^2(c+1)+c^2}{b^3+bc} \right]. \ \ \ \ \ (1)$ If $c = 1$ , then equation $(1)$ always holds, since both sides are $0$ . We obtain a family of solutions $(a, b) = (n, n^2+1)$ or $(a, b) = (n^2+1,n)$ . Note that the solution $(1, 2)$ found earlier is obtained for $n = 1$ . If $c>1$ , then equation $(1)$ implies that $\frac{b^2(c+1)+c^2}{b^3+bc}\ge (c-1)b$ . This simplifies to $c^2(b^2-1)+b^2(c(b^2-2)-(b^2+1)) \le 0. \ \ \ \ \ (2)$ Since $c \ge 2$ and $b^2-2 \ge 0$ , the only possibility is $b=2$ . But then equation $(2)$ becomes $3c^2+8c-20\le 0$ , which does not hold for $c\ge 2$ .

Hence the only solutions are $(n, n^2+1)$ and $(n^2+1, n)$ , $n \in N$ . And since for this problem $a<100$ answer will be $\sum_{n=1}^{99} (n^2+n+1) + \sum_{n=1}^{9} (n^2+n+1)=333738$