Solution of :$a^{10}b+1 = n^2$ with ($a,b,n \in \mathbb{N}$)

Question

I would like to find all couple $(a,b)$ such that :$$a^{10}b+1$$ is a perfect square yet I don't know how to proceed... Any ideas ?

Answer 1

First of all, it should be mentioned that, $a^{10}b=n^2-1$ or $a^{10}b=(n-1)(n+1)$. It has multiple solutions so I will provide some of them...

Solution #$0$:

In case of $a=1$ and $b=n^2-1$ the abovementioned equation is true for all natural $n$.

Solution #$1$:

You need to solve $n-1=b$ and $n+1=a^{10}$.

By this, I mean, $n=a^{10}-1$ and $b=a^{10}-2$. For example, $a=2, n=1023, b=1022$ is a solution.

Solution #$2$:

You need to solve $n-1=1$ and $n+1=a^{10}b$. This will give you $n=2, a=1, b=3$.

Solution #$3$:

You need to solve $n-1=ab$ and $n+1=a^{9}$. This will give you $n=a^{9}-1, b=a^{8}-\frac{2}{a}$. So put $a=1$ or $a=2$ . Then find values of $n$ and $b$.

Continue like this and you will find all of them...