What are the two smallest positive whole numbers the difference of whose squares is a cube and the difference of whose cubes is a square ?

(In reply to re(2): And after that there's... by Charlie)

I think the reason that multiplying both numbers by 64 renders more solutions is that 64=2^6 (though you probably already thought of that). Maybe multiplying both by 3^6 would work too. Maybe I can prove it for all u^6.

All of the following variables represent integers.
So, if x²-y²=z³ and x³-y³=t²,
would (x*u^6)²-(y*u^6)²= a perfect cube?
= x²*u^12-y²*u^12
= (x²-y²)*u^12
= a perfect cube!
Also, would (x*u^6)³-(y*u^6)³= a perfect square?
= (x³-y³)*u^18
= a perfect square!

So I was right. Why u^6 instead of u^5 or something else? Because 6 is a factor of both 2 and 3. I guess I was a little off topic there... it's not relevant to the puzzle.

When it says "the two smallest positive whole numbers," I think that that means we want to minimize the sum of the two numbers, though it could be interpreted in other ways. If you interpret as the smallest minimum of the two numbers, you could plausibly get a different answer.

Posted by Tristan on 2003-12-06 23:35:57