Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

It's highly unlikely that there is a way to factor $65(a^3b^3+a^2+b^2)-81(ab^3+1)=0$ and when we're asked to solve for the positive integer solutions, we would always think of treating the given equation as a quadratic function and then set its discriminant greater than or equal to zero, but this can't be the case here:

$65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$

If we're to rewrite it as a quadratic in $a$, we get:

$65a^3b^3 +65a^2 +65b^2 -81ab^3-81=0$

$65a^2(ab^3 +1)-81ab^3 +65b^2 -81=0$

Setting the discriminant $\ge 0$, we have:

$(-81b^3)^2-4(65)(ab^3 +1)(65b^2 -81)\ge 0$

That is a very messy inequality we've got there, and we should wise up to stop continuing and we should also know it wouldn't help if we treat the whole equation as a quadratic in $b$ either. That suggests we have to rewrite the given equation differently, how different, you may ask?

The answer is incredibly simple: As long as you can rewrite it in the form that you could deduce something wisely from it, then you're good.

Well, that is probably the answer that most students find to be extremely unhelpful! Okay, there actually has a rule of thumb that could guide us here, we need to rewrite the equation in such a way that we could create the situations where we only have to think of the range of, for example $a\le \text{some positive number, says k}$ and since we're told $a$ is positive integer, then we only need to check for the cases where $(1,\,2,\,\cdots,\,k)$ and we would then find for the corresponding $b$ integer solutions and we would be done.

Now, we take a look at my attempt:

I rewrite the equation as

$b^3(a)(65a^2-81)+65b^2+(65a^2-81)=0$ (*).

We have two cases to consider here, first is $65a^2-81>0$, which leads to $a≥2$. Replacing these two important information into (*) we see that we have the relation where it says always positive+always positive+always positive=0, which tells us there is no solution for $a≥2$.

That also means we only have to consider for $0\lt a≤1$. Now, solving the above equation for $b$ when $a=1$, we find $b=4$ and the only possible solution is $(a,\,b)=(1,\,4)$ we are hence done.