# Math Help - Solving for m and n

1. ## Solving for m and n

DO there exist positive integers m and n so that 76^m-34^n=10398457338?

I am looking for m and n that would solve this equation. I know that I need to work mod 5. So 76 is 1 (mod 5) and 34 is 4 (mod 5) and 10398457338 is 3 (mod 5) but how do i really solve for m and n,

DO there exist positive integers m and n so that 76^m-34^n=10398457338?

I am looking for m and n that would solve this equation. I know that I need to work mod 5. So 76 is 1 (mod 5) and 34 is 4 (mod 5) and 10398457338 is 3 (mod 5) but how do i really solve for m and n,
Note,
76=2*38
34=2*17
And m<n otherwise we have a negative number.

2^m*38^m-2^n*17^n=1039845338

Thus, the left hand side is divisible by 2^m
But the right hand side is divisible at most by 2 (not by 4=2^2).

Thus, this forces m=1.
In that case you can attempt to solve this equation for n since you determined what m has to be.

3. ## Right!!

I believe you are correct in saying this. I do not believe 34 to any power will give you that large number. I knew that 76 to the m power had to be 76 because of working in mod 5 will give 76 to the first power (mod 5) every time. I don't think that m and n can be found to solve this equation

4. Sorry to bring this up late after the discussion has ended, but I was reading through the forums and came across this problem.

I was a little confused by the proposed solution:

According to ThePerfectHacker, "m<n otherwise we have a negative number." I don't see how that's true.

If m<n and, according to the example ThePerfectHacker gave, m=1, then we get:
76^1-34^n=10398457338
76-34^n=10398457338
-34^n=10398457262 ... but that gives us a negative number.

I think it's more likely that n<m, in which case, using the same analysis:
2^m*38^m-2^n*17^n=1039845338

This is divisible by 2^n, but 1039845338 is only divisible by 2, therefore n=1.
76^m-34^1=10398457338
76^m-34=10398457338
76^m=10398457372

However, there is no value of m that gives this solution since 'the one's place' of any solution to 76^m where m is an element of the whole numbers will always be 6.
76^1=76
76^2=5776
76^3=438976
76^4=33362176

No solution can end in the number 2, such as 10398457372.

I should note that I'm not a student in Number Theory and have never taken the class so if I'm wrong in all the assumptions I've made, I apologize. I'm just curious to know if my assumptions are correct.
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