Peter wrote: Determine all pairs $ (a, b)$ of positive integers such that $ ab^2 + b + 7$ divides $ a^2 b + a + b$. $ ab^2+b+7|a^2b+a+b$ $ \Rightarrow ab^2+b+7|a^2b^2+ab+b^2$ $ \Rightarrow ab^2+b+7|a(ab^2+b+7)+b^2-7a$ $ \Leftrightarrow ab^2+b+7|b^2-7a$ Because $ ab^2+b+7>b^2-7a \rightarrow b^2=7a$ Let $ b=7k,a=7k^2$ Replace to the condition : $ 7^3k^4+7k+7|7^3k^5+7k^2+7k$ $ \Rightarrow 7^2k^4+k+1|k(7^2k^4+k+1)$ It is true for all $ k\in N$ So the solution is $ (a,b)=(7k^2,b=7k)$

TTsphn wrote: Peter wrote: Determine all pairs $ (a, b)$ of positive integers such that $ ab^2 + b + 7$ divides $ a^2 b + a + b$. $ ab^2 + b + 7|a^2b + a + b$ $ \Rightarrow ab^2 + b + 7|a^2b^2 + ab + b^2$ $ \Rightarrow ab^2 + b + 7|a(ab^2 + b + 7) + b^2 - 7a$ $ \Leftrightarrow ab^2 + b + 7|b^2 - 7a$ Because $ ab^2 + b + 7 > b^2 - 7a \rightarrow b^2 = 7a$ Let $ b = 7k,a = 7k^2$ Replace to the condition : $ 7^3k^4 + 7k + 7|7^3k^5 + 7k^2 + 7k$ $ \Rightarrow 7^2k^4 + k + 1|k(7^2k^4 + k + 1)$ It is true for all $ k\in N$ So the solution is $ (a,b) = (7k^2,b = 7k)$ This solution is incomplete; when $ b^2 - 7a < 0$, two solutions $ (11,1)$ and $ (49,1)$ are yielded. It is easy to prove that those two are the only ones satisfying the equation.