<s>[INST] How to prove that the roots of this equation are integers? Let there be an equation $a^2 + 4ab + b^2 - 121 = 0$ where I want to prove that a,b are integers. Then I want to find whether there are integer values of $b$ for which $a$ is also an integer. Let us consider the case for $a$ $\rightarrow$  $a = \dfrac{-4b \pm \sqrt{16b^2 - 4(b^2-121)}}{2} = -2b \pm \sqrt{3b^2 +121}$ Thereby the problem reduces to showing that $3b^2 +121$ is a perfect square for certain integer values of $b$. This is what I cannot do. 
Note: I know I can give examples and all for $3b^2 + 121$ to be a perfect square, but I'm looking for something with a little more substance. Thanks for the help.  
 [/INST] From $a^2+b^2+4ab-121=0$ we get $(a+b)^2+2ab=121$Or $2=\frac{11+a+b}{a} \frac{11-a-b}{b}$Now case 1: when the factors on right hand side are integers and coprime. Since 2 has only 2 factors, one of these has to be 1 and the other 2  Taking $\frac{11+a+b}{a}=2$ and $\frac{11-a-b}{b}=1$ we get $a=11 , b=0$
 Similarly $\frac{11+a+b}{a}=1$ and $\frac{11-a-b}{b}=2$ gives $a=44$ and $b=-11$
I can't do the case where those factors are not integers or coprime right now (note that's the case of $a=5,b=4$)
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