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 A powerful resemblance. (Posted on 2015-07-12)

1. Let a and b be consecutive non-zero whole numbers. Let C be the square of their sum, less their product, i.e. C = (a+b)^2-ab. Let n be a whole number.

Claim 1: C^n = (A+B)^2-AB, for some whole numbers A,B.

Prove it, or find a counter-example.

2. Let a and b be ANY relatively prime whole numbers, and perform the same operation.

Claim 2: C^n = (A+B)^2-AB, for some whole numbers A,B.

Any exceptions?

 No Solution Yet Submitted by broll Rating: 5.0000 (1 votes)

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 Stumped | Comment 1 of 2
I think this is a really neat problem, I am just completely stumped.

All I got was the n=2 subcase of claim 1: a=x+1, b=x, c=3x^2+3x+1, A=3n^2+2n, B=2n+1.  This subcase looks a lot like Searching for c among the cubes.

a=2, b=1, n=3 implies c=7. Then 7^3=343 has A=18 and B=1.  Where the B=1 comes from has me completely stumped.

 Posted by Brian Smith on 2017-01-01 18:27:04

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