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# Understand the problem

Find all integer solutions $(a,b)$ of the equation$\$(a+b+3)^2 + 2ab = 3ab(a+2)(b+2)\$$

##### Source of the problem
Costa Rica NMO 2010
Number Theory
6/10
##### Suggested Book
A Friendly Introduction to Number Theory by J.H.Silverman

Do you really need a hint? Try it first!

Write the equation in this form : $(a-b)^2 = 3(ab-1)(ab+2a+2b+3)$ The idea is that the RHS is a 4 degree polynomial in two variables and the LHS is a 2 degree polynomial in the same two variables. Now, you have the intuition that the 4th-degree polynomial will surpass the LHS after some time. We, therefore, aim to bound the a and b assuming the equality holds.

So, we try to find when RHS $\geq$ LHS.  Rather we will find the condition when $(a-b)^2 \leq (ab-1)(ab + 2a + 2b + 3)$. Assume $a \geq b$. We will find when $latex (a-b) \leq (ab-1)$ and $(a-b) \leq (ab + 2a + 2b + 3)$.

$latex (a-b) \leq (ab-1)$ – For this to hold. $(a^2-1)(b^2 -1) \geq 0$. $(a-b) \leq (ab + 2a + 2b + 3)$ – For this to hold $(a+3)(b+1) \geq 0$.

Hence, try to understand that we have essentially bounded the solutions. Observe this implies that to solutions to have essentially. $latex -3 \leq a,b \leq 1$. Hence show that by computation that : The solution set is  $\boxed{(-3,-3), (-3,0), (0,-3), (-2,1), (1,-2), (-1,-1), (1,1)}$

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