Question

prove that 2 power n + 6 into 9 power n is divisible by 7 for any Integer n by using congruence modulo​

Answer 1

Step-by-step explanation:

So you have n6≡1(mod7), but this implies n7≡n(mod7). So 7|n7−n. In general, np≡n(modp) for a prime p. By the same token, n7≡(n3)2⋅n≡n3≡n(mod3) by FlT, so 3|n7−n. It remains to show 2|n7−n,

Answer 2

Answer:

So you have n6≡1(mod7), but this implies n7≡n(mod7).

So 7|n7−n. In general, np≡n(modp) for a prime p.

By the same token, n7≡(n3)2⋅n≡n3≡n(mod3) by FlT, so 3|n7−n. It remains to show 2|n7−n,