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Find From Four 2 (Posted on 2016-06-23) Difficulty: 3 of 5
Each of X and Y is a positive integer such that:

X4 and Y4 share identical last four digits in the same order, and

X-Y = 2016

(A) Find the smallest solution satisfying the given conditions.

(B) Derive the general form of X and Y satisfying the given conditions.

No Solution Yet Submitted by K Sengupta    
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A way | Comment 1 of 4
X^4-Y^4 congruent 0 (mod 10^4)

As Y=X-2016 and reordering
X^4-Y^4=8064 (X-1008) (X^2-2016X+2032128)

8064*625 is the lower 0 (mod 10^4) I've find as a product of 8064.
 So if (X-1008) is multiple of 625 we obtain solutions

For X=2258, X-1008=625*2. Then Y=242

X=2258     X^4=25995354862096
Y=242       Y^4=34297242096

Then we know that for X=2258+625n  Y=242+625n there will be solutions. 

Edited on June 23, 2016, 4:56 pm
  Posted by armando on 2016-06-23 16:15:47

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