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,
Originally Posted by chadlyter
And m<n otherwise we have a negative number.
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.
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
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:
-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:
This is divisible by 2^n, but 1039845338 is only divisible by 2, therefore n=1.
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.
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.