Little math problems September 2018

[pier4r]

1) albert commenting his own post.

2) Sorry I thought it was not needed. I find all the effort put into solutions neat. Of course even further explorations are appreciated. I find that good(great/outstanding/etc..) contributions are good even without comments. A good contribution is not good only when commented.

3) I guess I claimed too early something I (and the other guys at the math jam) didn't test enough. At the math jam I observed (the other guy did it too independently and then we confronted the solutions n1) the following:

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2^2 + 2^2 = 8 = 2^3
2^3 + 2^3 = 16 = 2^4 = (2^2)^2

2^8 + 2^8 = 2*2^8 = 2^9 = (2^3)^3 
2^9 + 2^9 = 2*2^9 = 2^10 = (2^5)^2 

so we quickly settled on (we didn't test much) odd powers of two as exponent.

2^32 + 2^32 = 2*2^32 = 2^33 = (2^11)^3
2^33 + 2^33 = 2*2^33 = 2^34 = (2^17)^2 

2^128 + 2^128 = 2*2^128 = 2^129 = (2^43)^3
2^129 + 2^129 = 2*2^129 = 2^130 = (2^65)^2

2^512 + 2^512 = 2*2^512 = 2^513 = (2^171)^3
2^513 + 2^513 = 2*2^513 = 2^514 = (2^257)^2

2^(2^(2k+1)) + 2^(2^(2k+1)) = 2*2^(2^(2k+1)) = 2^(2^(2k+1)+1)
Now indeed we didn't prove that 2^(2k+1)+1 is always a multiple of 3.
2^(2^(2k+1)+1) + 2^(2^(2k+1)+1) = 2*2^(2^(2k+1)+1) = 2^(2^(2k+1)+2)
While 2^(2k+1)+2 , still we didn't really check at the time, is a multiple of 2.

If 2^(2k+1)+1 is a multiple of 3 always, then there is a family of solutions. I will check later.

n1: the idea at the math jam is to work on a problem on your own, then share the solutions and double check them. If everyone is stuck, then brainstorming together.

update: what the solutions in this threads shows is that there seems to be other fitting numbers as well. We had only pen + paper in a bar at the mathjam so we didn't stay on the problem too long.