Let first $ a = b > 1$. Then $ a^{2} - a\mid a^{2} + a$ implies $ a^{2} - a\mid 2a$; hence $ a^{2} - a\leq 2a$ which is only possible for $ a\in\{2,3\}$. Simple checking shows that $ a = 2$ and $ a = 3$ are indeed solutions. Let's now assume that $ a\neq b$. Wlog assume that $ a > b$. Then $ a^{2} - b\mid b^{2} + a$ implies $ a^{2} - b\leq b^{2} + a$, so $ a(a - 1)\leq b(b + 1)$. Since $ a > b$, this is only possible for $ a = b + 1$ (in this case, we have equality). Now, $ b^{2} - a\mid a^{2} + b$ implies $ b^{2} - b - 1\mid b^{2} + 3b + 1$ and thus $ b^{2} - b - 1\mid 4b + 2$. This is true for $ b = 1$, for $ b\geq 2$, we have $ b^{2} - b - 1 > 0$ and hence we need $ b^{2} - b - 1\leq 4b + 2$, which is only possible for $ b\leq 5$. Simple checking shows that only $ b = 2$ and $ a = 3$ is actually a solution. We hence get the solutions \[ (a,b)\in \{ (2,2); (3,3); (1,2); (2,1); (2,3); (3;2)\}. \]

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