Mathematical Reflections 2013, Issue 4 - Problem J276

Problem:
Find all positive integers $m$ and $n$ such that $$10^n-6^m=4n^2$$

Proposed by Tigran Akopyan.

Solution:
It is easy to see that if $n=1$, then $m=1$ and $(1,1)$ is a solution to the given equation. We prove that this is the only solution. Assume that $n>1$ and $n$ odd. Then, $10^n-6^m \equiv -6^m \pmod{8}$ and $4n^2 \equiv 4 \pmod{8}$ and it is clear that $-6^m \equiv 4 \pmod{8}$ if and only if $m=2$. But $10^n>36+4n^2$ for all positive integers $n>1$, therefore there are no solutions when $n$ is odd. Let $n$ be an even number, i.e. $n=2k$ for some $k \in \mathbb{Z}^+$. Hence,
$$(10^k-4k)(10^k+4k)=6^m.$$ Since there are no solutions for $k=1,2$, let us assume that $k>2$. Clearly $m \geq 4$ and simplifying by $2$, we get
$$\begin{equation} (2^{k-2}\cdot5^k - k)(2^{k-2}\cdot5^k + k)=2^{m-4}\cdot3^m.                          (1) \end{equation}$$
If $k$ is odd, then $m=4$. But $2^{k-2}\cdot5^k + k \geq 2\cdot5^3+3 > 3^4$, contradiction. Therefore, $k$ must be even. Assume that $k=2^{\alpha}h$, where $\alpha,h \in \mathbb{Z}^+$ and $h$ is odd. If $\alpha \geq k-2$, then $2^{k-2} | k$, which implies that $2^{k-2} \leq k$. This gives $k=3,4$ and an easy check shows that there are no solutions for this values. If $\alpha < k-2$, then equation $(1)$ becomes $$2^{2\alpha}(2^{k-2-\alpha}\cdot5^k - h)(2^{k-2-\alpha}\cdot5^k + h)=2^{m-4}\cdot3^m,$$ and by unique factorizaton we have $2\alpha=m-4$. Therefore, from the inequality $k>\alpha+2$, we obtain $n>2\alpha+4=m$. But this implies that $10^n=6^m+4n^2<6^n+4n^2$, which is false for all integers $n>1$. Hence, there are no solutions for $n$ even and the statement follows.