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Seven Divisibility Settlement (Posted on 2010-11-13) Difficulty: 3 of 5
M is a positive integer ≥ 2 such that 3M + 4M is divisible by M.

Is M always divisible by 7?

If so, prove it. Otherwise, provide a counterexample.

No Solution Yet Submitted by K Sengupta    
Rating: 2.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Possible Solution | Comment 2 of 3 |
(In reply to Possible Solution by broll)

I don't get why only prime M need to be checked.


Take M=35 for example.  Fermat's little theorem does indeed show 3^35 mod 7 = 3 and 3^35 mod 7 = 3, but 3^35 mod 35 = 12.  Also 4^35 mod 5 = 4 and 4^35 mod 4 = 4, but 4^35 mod 35 = 9.

I think all that you proved is 7 is the only prime M that divides 3^M+4^M.

Before I hit an overflow, direct calculation finds M=1, 7, 49=7^2, 343=7^3, 2401=7^4, and 2653=7*379 are values that divide 3^M+4^M.  Going to the OEIS finds sequence A045584 which other than 1, are all multiples of 7.

Edited on June 24, 2017, 1:54 pm
  Posted by Brian Smith on 2017-06-24 13:44:34

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