Arne Smeets

Posted on Tuesday, 09 December, 2003 - 05:43 pm:

Prove that for each positive integer k, there is a positive integer m such that 2^k |3^m +5.
Looking for an elegant solution ...

David Loeffler

Posted on Wednesday, 10 December, 2003 - 11:44 pm:

Here's a random idea. It's easily seen by induction that 3^{2^k-3}=2^k-1+1 mod 2^k, so 3 has order 2^k-2 in the multiplicative group. Also, -1 is not a power of 3 mod 2^k, since it would have to be 3^{2^k-3}and it isn't.

It's now obvious that exactly one of 5 and -5 is a power of 3. That's almost there, but I don't quite see how to finish it off from here. Anyone got any ideas?

David

Kerwin Hui

Posted on Thursday, 11 December, 2003 - 01:54 pm:

David, mod 8 finishes off the proof quite nicely.

Kerwin

David Loeffler

Posted on Thursday, 11 December, 2003 - 10:25 pm:

Indeed - that occurred to me about 10 minutes after I logged out and shut down my PC last night! I should perhaps point out that my statement about the order of 3 in the multiplicative group only works for k