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Really simple problem which has gone over my head watch

1. Hello again, today I got stuck on another problem which looked very simple but even after nearly an hour of manipulation got me nowhere nearer the answer except for telling me what the answer was not I decided to ask for help.
So here is the problem:

Find positive integers and such that

As you can see this problem looks deceptively simple however it is anything but. If anyone has any hints then please post them.

2. OK, this isn't nice, but it works.

Writing A, B for the cube roots, the LHS = (A+B-1)^2 = A^2 + B^2 + 2(AB - A - B) + 1

Assuming A, B aren't integers, the only place an integer can come from is the 2AB term.

So we must have 2AB+1 = 49, or AB = 24. So ab = (24)^3 = (3 x 2^3)^3 = 3^3 2^9.

So assume a = 2^p 3^q (and so b = 2^(9-p) 3^(3-q)). Wlog, p < 5 (since otherwise can swap b, a to get p < 5). And it's clear neither p, q = 0.

So 1<=p<=4, 1<=q<=2. At this point trial and error should suffice.
3. (Original post by EmptyMathsBox)
..
(quoted for visibility)

Incidentally, I cheated to solve this (computer program) and then reverse engineered the algebraic solution. It turns out that a, b are much larger than my intuition expected - definitely not one to do by basic trial and error manually!
4. Thank you very much for the answer. However I am wondering as to how you worked out that only 2AB could be an integer. Why could not be an integer (if a is a perfect cube)?
5. (Original post by EmptyMathsBox)
Thank you very much for the answer. However I am wondering as to how you worked out that only 2AB could be an integer. Why could not be an integer (if a is a perfect cube)?
Well, if a was a perfect cube, it wouldn't be an interesting question. (If you look at what I wrote, I assumed A, B weren't integers, and this was my reasoning for that).

Obviously that's not a proof - for a formal solution you'd need to rule it out. But there aren't many possible cubes to consider (before the LHS becomes too big for the RHS).

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