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(n3+1)/(mn-1)

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Posted · Report post

Find all pairs of positive integers m and n such that (n3+1) / (mn-1) is an integer, and prove you've identified them all.

This was a problem I solved back in high school with a proof that could barely fit on one page, written front and back, with small handwriting.

Even looking at it again now, the most elegant proof I can come up with is still pretty complex, but nevertheless sort of neat.

In part, I'm sharing a tricky problem. In part, I'm wondering if there's a more elegant way of solving it than the one I came up with.

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Posted · Report post

n = 1, m = 3
n = 2, m = 2, m = 5
n = 5, m = 2

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Posted · Report post

n, m

1, 2

1, 3

2, 1

2, 2

2, 5

3, 1

3, 2

3, 5

5, 2

5, 3

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Posted · Report post

I don't have the time to put together a complete proof, but here is how I would go about it...

Show that n3+1 = (n+1)(n2-n+1) and this is the only way to represent it as a product. Therefore for (n3+1)/(mn-1) to be an integer one of the following conditions must be true for some integer k>0:

1) n+1 = k(mn-1)

2) n2-n+1 = k(mn-1)

For each of these cases show that there exist small values of m and n, for which these equations are true, but also show that for larger m and/or n the left and right side of these equations diverge and no more solutions are possible.

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Posted · Report post

I'll say that that approach would certainly be a different way of going about it than what I did.

Would it matter that 27 is not divisible by 21, and 49 is not divisible by 21, but 27*49=1323 is divisible by 21 (=21*63)?

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Posted · Report post

:duh: So much for a "proof in a hurry" LOL

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