Find all positive integers s.t. $10^m-8^n=2m^2$

Question

Find all pairs of positive integers $(m,n)$ such that $10^m-8^n=2m^2$

Answer 1

Hint:

$${10}^m - 8^n = {2^m}{5^m} - {2^{3n}}$$

so that

$$m^2 = {2^{m-1}}{5^m} - {2^{3n-1}}.$$

In particular:

$${2^{m-1}}{5^m} - {2^{3n-1}} \geq 1,$$

and

$${2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$

is a square, which means $m \equiv 1 \pmod 2$.

Suppose that $m > 1$. Then the RHS of

$$m^2 = {2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$

is even while the LHS is odd (since $5^m - {2^{3n-m}}$ is always odd, unless $m = 3n$ [I'll deal with this case later]). A contradiction. Therefore, $m = 1$.

Substituting $m = 1$:

$$10 - {8^n} = 2$$

which gives $n = 1$.

Update: If $m = 3n$, then

$$m^2 = {3^2}{n^2} = 2^{3n - 1}\left(5^{3n} - 1\right).$$

Since $\gcd(3,2) = 1$, then it follows that

$$2^{3n - 1} \mid n^2$$

which implies that $2^{3n - 1} \leq n^2$, contradicting $n^2 < 2^{3n - 1}$.

Therefore, in general, $m = 3n$ cannot occur.

Hence the only solution to the original problem is $(m, n) = (1, 1)$.