Find all integer numbers $k$,$l$,$m$,$n$ for which $3^k$ • $5^l$ + $7^m$ = $2^n$

Question

Find all integer numbers $k$,$l$,$m$,$n$ for which $3^k\cdot 5^l+7^m=2^n$.

I tried to find a decent modulo and look at ending digits, but nothing more. Thank you for your responses. Obviously is $k$ = $l$ = $m$ = $0$ and $n$ = $1$ a solution.

Answer 1

From Robert's list: there is an elementary method to find all solutions to, say, $$ 2^n - 7^m = 15. $$ Evidently we need to solve the same problem when replacing $15$ by these $1,3,5,9,15, 25.$ I do not see any reason to expect a proof that $k+l \leq 2.$

For the six problems mentioned, see

Exponential Diophantine equation $7^y + 2 = 3^x$

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Finding solutions to the diophantine equation $7^a=3^b+100$

Gottfried Helms came up with a different algorithm that allowed him to solve problems where my numbers got too big.

Answer 2

I find the following values: $$ \matrix{ k & l & m & n\cr0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 3 \cr 0 & 2 & 1 & 5 \cr 1 & 0 & 0 & 2 \cr 1 & 1 & 0 & 4 \cr 1 & 1 & 2 & 6 \cr 2 & 0 & 1 & 4 \cr }$$ These are the only ones with $0 \le k,l,m \le 100$.

Answer 3

let (1): $3^k 5^l+7^m=2^n$ (supposing that k and l are greater or equal to 1). Studying the equation modulo 3, you see that n is even, $n=2n'$. Studying now the equation at modulo 5, you see too that $n-m = 4a$, for some $a$. Pass the 7^m to the right, factorize and observe that $2^{n'}-7^{2a-n'}$ and $2^{n'}+7^{2a-n'}$ are coprime. The rest its easy and you can finish without much dificulty.