Question 499469: find the pairs of m and n such that m!+n!=m(to the power of n)
Found 2 solutions by chessace, richard1234:
Answer by chessace(471)   (Show Source):
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! (m,n) = (1,0), (2,2), and (2,3) work (yes, 0! = 1).
Suppose that m and n are both larger than 2. Since all factorials >= 3! are divisible by 6, it follows that m! + n! is divisible by 6, so the RHS is also divisible by 6. This only holds when m is divisible by 6. In addition, since we fixed n >= 3, then the RHS is divisible by 216, so the LHS, similarly, is divisible by 216.
Since m is divisible by 6, then m must be at least 6, so m! is at least 720 (congruent to 72 mod 216). If m = 6, then n = 8 (so that this works modulo 216), and if m is greater than 6 such that m! ≡ 0 (mod 216), then n can be large enough. It will follow that the LHS and RHS will get extremely large, in which certain restraints from one side of the equation will restrain the other side even more. If this logic applies, then there can be no solutions (I hope).
The exception is if one of m, n is 1 or 2. If m = 2, then n cannot be larger than 3 (otherwise LHS ≡ 2 (mod 4)). (2,3) works here. If n = 2, we have m! + 2 = m^2; doesn't work for large m.
Setting m or n equal to 1 will eventually fail.
So it seems like the only solutions are (1,0), (2,2) and (2,3). I highly doubt any other solutions exist.
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