Is there any solution to this Diophantine equation?

Question

Consider the equation $a^n+n=b^2$, where $(a,n)>2$ are positive integers. Is it true that there exist no integer solutions $b$ of this equation? Can one find a counterexample?

I am unable to do either.

Answer 1

The Diophantine equation $$ y^n=f(x) $$ for an irreducible polynomial $f$ with integer coefficients of degree $m\ge 2$ has only finitely many solutions, provided $(m,n)\neq (2,2)$. This is due to Carl Ludwig Siegel. The following reference discusses the special case $f(x)=x^2+c$, which might be interesting.

Reference: see here.

Answer 2

This is also an extended comment

There are some bounds on sum of logs like Baker/Rhin studied, but I don't know if it can be generalized.

This would go like this:

$$\frac{b^2}{a^n}-1>\log(\frac{b^2}{a^n})=|2\cdot\log(b)-n\cdot\log(a)|$$

Using the bound $|µ_1\log (b)+µ_2\log (a)|\geq H^{-c}$ with $H=max(|µ_1|,|µ_2|)$) we would have

$$|2\cdot\log(b)-n\cdot\log(a)|>\frac 1{n^{c}}$$ $$\frac{b^2-a^n}{a^n}>\frac 1{n^{c}}$$ $$n=b^2-a^n>\frac {a^n}{n^{c}}$$ $$n^{c+1}>a^n$$

Knowing that $a>2$ and $c$ can be relatively small, the size of $n$ in the last inequality must also be very limited.