Diophantine equation in the integers

Question

Find all integers $a,b$ such that

$6(a^2-ab+b^2) = 31(a+b)$

Ideas I have had so far:

• Move everything to the left side of the equation and try to solve a polynomial in $a$,whose discriminant in $b$ is $-108b^2 + 1116b + 961$.Now I cannot see how this will lead anywhere.
• The equation can be written as: $6[(a+b)^2 - 3ab] = 31(a+b)$. If we set $a+b=s$ and $ab = p$, we get that $6(s^2-3p) = 31s$. Solving to get p we see that $p = \frac{6s^2-31s}{18}$ and so $18$ must divide $6s^2-31s$. I like where this is going but I cannot move from there.

Any helpful ideas or hints?

Answer 1

Using the inequality: $ab \leq \dfrac{(a+b)^2}{4}$, we have: $\dfrac{6s^2-31s}{18} = p = ab \leq \dfrac{(a+b)^2}{4} = \dfrac{s^2}{4}\Rightarrow 24s^2 - 124s \leq 18s^2 \Rightarrow 6s^2 - 124s \leq 0 \Rightarrow s(6s - 124) \leq 0 \Rightarrow 0 \leq s \leq \dfrac{124}{6} < 22$. You can take it from here.