2^m = 3^n +5

[PrudensOptimus]

Hello,


2^m = 3^n + 5, given m,n >=0

... Find all possible m,n... Any ideas?

[Hurkyl]

Try looking at it in varios moduli?

[PrudensOptimus]

what does that mean lol pls explain.

[Hurkyl]

For example, if you reduce the equation mod 2^m, and there are no values of n such that 3^n + 5 = 0 (mod 2^m), then you've found an upper limit for m.

I don't know if this will work...

[Tide]

Try rewriting your equation as
2^m - 2 = 3^n + 3
That might provide some insight.

[robert Ihnot]

Well, two small solutions are: 2^3 = 3+5; 2^5=3^3+5.

[devious_]

Try looking up Stroeker and Tijdeman's theorem (a.k.a. the solution to Pillai's conjecture).