Divisible by 2004

[BetonaBG]

First of all, I'd like to say Hello to everyone as this is my first post :)

Now the fun part, I spend 2-3h in painful strugle and I was unable
to prove the following question. If anyone knows how to solve it
it will be largely appreciated :)


Prove for any positive integer n that:
2196^n – 25^n – 180^n + 13^n is divisible by 2004



Have Fun

[haggy]

2004 = 167*3*4
These are relatively prime to one another.
2196^n – 25^n – 180^n + 13^n = (13*167+25)^n -25^n -(167+13)^n +13^n
Therefore
2196^n – 25^n – 180^n + 13^n = 25^n -25^n -13^n +13^n (mod 167)
2196^n – 25^n – 180^n + 13^n = 0 (mod 167)
Likewise
2196^n – 25^n – 180^n + 13^n = 0 (mod 3)
2196^n – 25^n – 180^n + 13^n = 0 (mod 4)

Therefore
2196^n – 25^n – 180^n + 13^n = 0 (mod 2004)

[BetonaBG]

I'll say simple - Thanks


but my gratitude pales in comparatively with my words.