\(x\) and \(y\) are integers such that \(3x^2y^2 - 17xy + 20 = 0.\) What is the maximum possible value of \(x+y?\)
+6248 \(\text{Notice the expression is symmetrical in $x$ and $y$}\\ \text{This suggests that any solution will be such that $x=y$}\\ $\text{Letting $x=y$ we have}\\ 3x^4 - 17x^2 + 20 = 0\\ \text{let $u=x^2$}\\ 3u^2 - 17u + 20 = 0\\ \text{Apply the quadratic formula}\\ u = \dfrac{17 \pm \sqrt{17^2 - (4)(3)(20)}}{6} = \left(\dfrac 5 3,~4\right)\\ \text{you want integer solutions so we can ignore the first one}\\ x^2 = 4\\ x=\pm 2\)
\(\text{So the max of $x+y$ is $2+2=4$}\)
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