As Charlie posts a complete pattern of the solutions, I can try to give a more complete answer to my precedent post.
X^4-Y^4 is congruent 0 (mod 10^4)
As Y=X-2016 its possible to express the differnce in terms of X. Doing it and reordering
X^4-Y^4=8064 (X-1008) ((X-1008)^2+ 1008^2)
8064*625 is the lower 0 (mod 10^4) I've find as a product of 8064. And as 10000=625*16, numbers congruent to mod 10^4 will also be congruent to mod 625.
When the expression in bold is multiple of 625 there will be solutions.
This happens if:
- (X-1008) is multiple of 625, or if
- ((X-1008)^2+ 1008^2) is multiple of 625, or if
- neither of both expressions is multiple of 625 but the product of both is.
But this third case is not possible.
625=5^4, so if it is the product of two numbers both have to be multiple of 5. But when (x-1008) is multiple of 5, then (x-1008)^2 will also be, and then ((X-1008)^2+ 1008^2) won't be, because 1008^2 is not.
The third case being impossible we are left with the first two cases:
a) The first set of solutions is when X-1008 is multiple of 625. Here is easy to calculate X: X=625n+383 (383 being the congruence of 1008 to mod 625).
The lowest solution for that set is X=2258 Y=242
b) There will be other sets of solutions if ((X-1008)^2+ 1008^2) is multiple of 625.
For this to happen: as 1008^2 is congruent with 439 (mod 625), necessarily (X-1008)^2 should be congruent to 625-439=186 (mod 625).
It means that (X-1008)^2/625 would have as decimal part 186/625=0,2976.
From this: ¿how may I know X-1008? I'don't know.
Anyway there are only 625 possibilities, which is not too much for Excel. We can calculate one intere cycle of the expression
(X-1008)^2/625 varying X from 1008 to 1634 and check when the decimal part is ,2976.
It comes out that for X-1008=294 and 331 the expression has as decimal part ,2976. From here we obtain X and Y. X will be congruent to 294+383=677, so to 52 (mod 625), etc.
Then there are also solutions if:
X=625n+52 (Y=625n - 89) Lowest X=2552 Y=536
X=625n+89 (Y=625n - 52) Lowest X=2589 Y=573
So when X is congruent to 52, 89, 383 (mod 625) there are solutions to the puzzle. (Is the same solution Charlie expressed in bold terms for Y).
Edited on June 25, 2016, 5:10 am
Posted by armando
on 2016-06-25 04:20:09