Division of Powers

Show that 2903n - 803n - 464n + 261n is divisible by 1897 for all n>0.

Can you find an example of this for smaller values?
A small one I found is: 8n + 7n - 3n - 2n is divisible by 10 for all n>0.

Source: Reader Jimmy Chng Gim Hong, from Singapore.

Solutions were received from Frank Nestel, Radu-Florian Atanasiu (from Romania), Paul McMahon, Ken Suprin, Radu Ionescu, Jimmy Chng Gim Hong, Sudipta Das, Lou Cairoli, Philippe Fondanaiche, Joseph DeVincentis, John Hewson, Brendan Owen, and Ross Millikan. There were a variety of solutions. Here are a few.
A general statement from Radu Ionescu (without proof):
a^n -b^n + c^n - d^n is divisible by GCD((a-b),(c-d))* GCD((a-d),(c-b)) for all n>0.
Ross Millikan used the Chinese Remainder Theorem (without actually mentioning it):
As 1897 is 7*271, we should consider this mod 7 and mod 271. 2903 is 5 mod 7, -803 is -5 mod 7, -464 is -2 mod 7, and 261 is +2 mod 7. So mod 7, we have 5^n - (5)^n - (2)^n + 2^n = 0. Similarly mod 271 we have -(78)^n + 10^n + 78^n - (10)^n.
Most people simply proved the sum was divisible by 271 and 7 using the method similar to Radu-Florian Atanasiu: 
    Because 1897 = 271 * 7 and (271,7) = 1, it is enough to verify  the
dividing of the expression E = 2903^n - 803^n - 464^n + 261 ^n by 271 and
then by 7.  We will use the statement that for any n natural,

        (a-b) | (a^n - b^n) , for any (a <>b).

    Then:

 E = (2903^n - 464^n) - (803^n - 261^n) = (2903 - 464)A - (803 - 261)B =
2439A - 542B = 271(9A - 2B);

 E = (2903^n - 803^n) - (464^n - 261^n) = (2903 - 803)A - (464 - 261)B =
2100A - 203B = 7(300A - 29B).

   And so our problem is solved.
Some smaller examples people found:
Mail to Ken