# prove that , A=(2903)^n - (803)^n - (464)^n +(261)^n , is divisible by 1897.

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### 1 Answer

To prove that , A=(2903)^n - (803)^n - (464)^n +(261)^n , is divisible by 1897.

Proof:

2903 = 193+271*10. So 2903 = 193 (mod271).

|||ly, we see that

803 = 261 (mod 271)

464 = 193 (mod271)

261 = 261 (mod271).

Therefore A = (2903)^n - (803)^n - (464)^n +(261)^n =>

=> A = 193^n(mod 271) - 261^n(mod 271) -193^n(mod271)+261(271).

=> A = 0 (mod 271).

=> 271| A. Or 271 divides A ....(1)

Again we express we see that:

2903 = 55 (mod 7)

803 = 5 (mod 7)

464 = 2 (mod 7)

261 = 2 (mod 7).

Therefore, A=(2903)^n - (803)^n - (464)^n +(261)^n =>

=> A = 5^n(mod7) - 5^n (mod7) - 2^n(mod7) +2^n (mod7)

=> A = 0 (mod 7)

=> 7| A,. Or & divides A.

From (1) and (2) we see that 7 and 271 both divide the number A=(2903)^n - (803)^n - (464)^n +(261)^n .

7*271 = 1897.

**Therefore the number 1897 divides the number A=(2903)^n - (803)^n - (464)^n +(261)^n** for all integers n.