Squares and Cubes

[Sriman Dutta]

Hi,

Can anybody find two numbers such that the difference between their squares is a cube and the difference between their cubes is a square ? :-)

[DrKrettin]

Am I on the right track if I take the two numbers to be 2^n plus and minus 1? If n is a multiple of 3, then at least I get the first condition.

[Sriman Dutta]

There must be two numbers a and b such that a^2 - b^2 = n^3 and a^3 - b^3 = m^2 , m and n are two different numbers.

[DrKrettin]

Yes, I got that. I meant whether a good starting point might be taking a = 2^n - 1 and b = 2^n + 1 where n is a positive integer. Then a^2 - b^2 = (a + b)(a - b) = 2^(n+2) which is a cube for n= 1,4,7..

[Sriman Dutta]

OK

And the other condition ?

[uncool]

DrKrettin: that wouldn't work, as then a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 2*something odd, which can't be a square (except in the case that n = 0, a = 2, b = 0, and it still isn't a square).

Edited October 21, 2016 by uncool

[DrKrettin]

DrKrettin: that wouldn't work, as then a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 2*something odd, which can't be a square (except in the case that n = 0, a = 2, b = 0, and it still isn't a square).

 

Yes - I'd decided that approach was getting nowhere. There doesn't seem to be an analytical way of doing this.

[imatfaal]

10 and 6

[DrKrettin]

Nice. Do you know whether that is a unique answer, or are there other larger ones?

[imatfaal]

Nice. Do you know whether that is a unique answer, or are there other larger ones?

 

No idea - ran through a few numbers in my head on my ride to work today. I cannot think of an analytical method - or any way other than sieve tbh

[Sriman Dutta]

Perfect imatfaal and +1