Little math problems September 2018

[Albert Chan]

(09-24-2018 12:53 AM)Valentin Albillo Wrote:        (2,2), (128, 128), (1250, 625), (1458, 1458), (8192, 8192), (31250, 31250), (80000, 40000)

and infinitely more can be found by simply expanding the search range, though at this point it's more efficient
(albeit more convoluted) to begin using congruences and parameterization, as stated.

Is there a pattern to the solution pairs ?
For above, it seems to be (2 N^3, 2 N^3) or (2 N^3, N^3).

Quote:Another way to proceed is to perform algebraic manipulations using divisibility criteria ...

What divisibility criteria did you end up using ?