Can $10^n+89$ ever be a Perfect Square for $n>3$? Is it correct that if an natural number $10^n+89$ is a perfect square then $n=3$?
The answer is clear if $n$ is an even number. For odd $n$ I can prove that $10^n+89$ can be a perfect square only if $n=22m+3$...
 A: This is a semi-proof that there are only finitely many solutions. I use the abc-conjecture here, which has so far not been confirmed to be proven (although Shinichi Mochizuki claimes to have a proof, hence my "semi-proof").


The abc-conjecture.
Given $a,b,c>0$ with $a+b=c$ and $\gcd(a,b,c)=1$, for every $\epsilon>0$ there exists a $K_\epsilon$ such that
  $$c<K_\epsilon\operatorname{rad}(abc)^{1+\epsilon}$$
  where $\operatorname{rad}(x)$ denotes the product of the distinct prime factors of $x$.

So given that $10^n+89=s^2$, we know that for every $\epsilon>0$ there exists a $K_\epsilon$ such that
$$s^2<K_\epsilon\operatorname{rad}(10^n\cdot 89\cdot s^2)^{1+\epsilon}$$
note that $\operatorname{rad}(abc)=\operatorname{rad}(a)\operatorname{rad}(b)\operatorname{rad}(c)$ (for pairwise coprime $a,b,c$, and notice that $10^n$, $89$ and $s^2$ are pairwise coprime). So
\begin{align}
s^2&<K_\epsilon\operatorname{rad}(10^n\cdot 89\cdot s^2)\\
&=K_\epsilon\left(\operatorname{rad}(10^n)\operatorname{rad}(89)\operatorname{rad}(s^2)\right)^{1+\epsilon}\\
&=K_\epsilon\cdot\left(10\cdot 89\cdot \operatorname{rad}(s)\right)^{1+\epsilon}\\
&=K_\epsilon\cdot890^{1+\epsilon}\operatorname{rad}(s)^{1+\epsilon}\\
&\leq K_\epsilon\cdot890^{1+\epsilon}s^{1+\epsilon}
\end{align}
which means that
$$s^2<K_\epsilon\cdot890^{1+\epsilon}s^{1+\epsilon}$$
and dividing by $s^{1+\epsilon}>0$ gives
$$s^{1-\epsilon}<K_\epsilon\cdot890^{1+\epsilon}$$
for a fixed $\epsilon>0$, $K_\epsilon$ is fixed, and so is $890^{1+\epsilon}$, so there are at most finitely many solutions for $s$ (since $s$ is positive and bounded above by $\sqrt[1-\epsilon]{K_\epsilon890^{1+\epsilon}}$). Unfortunately, we don't know what $K_\epsilon$ is (except the lower bound $K_\epsilon>\frac{33}{890}29370^{-\epsilon}$ which we can find using the only solution we have), so unfortunately we still can't rule out those finitely many posibilities to get only $s=33$ (that's the solution $n=3$). 
