find the minimum value of $N$ given that $N$ is a positive integer and $N=3a^2-ab^2-2b-4$ for positive integers $a$ and $b$

Question

Question:

Find the minimum value of $N$ given that $N$ is a positive integer and $N=3a^2-ab^2-2b-4$ for positive integers $a$ and $b$

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My attempt:

By a bit of guessing and checking I was able to obtain $N=2$ for $a=4$ and $b=3$ but I am unable to prove that $N$ is not equal to $1$ therefore I am looking for a much cleaner solution.

Answer 1

Suppose for some $a, b$ we have $N = 1$. That is, $3a^2 - b^2a - (2b + 5) = 0$, viewed as a quadratic equation of $a$.

Taking determinant, we get $b^4 + 24b + 60$, which must be a perfect square. Since it's strictly larger than $(b^2)^2$, it must be at least $(b^2 + 1)^2$.

Thus we have $(b^2 + 1)^2 \le b^4 + 24b + 60$, which leads to $2b^2 - 24b - 59 \le 0$, and therefore $b \le 14$.

Checking all $b$ from $1$ to $14$, none of them gives $b^4 + 24b + 60$ a perfect square. Therefore no positive integers $(a, b)$ would give $N = 1$.