Prove this...?

Question

that seven divides 5^(2n)+3*2^(5n-2)
dont use induction
use congruence

Answer 1

Recall the following basic facts:

5^(2n)=25^n and 2^(5n-2) = 2^(-2)*(2^5)^n = 4^(-1)*32^n.

Working modulo 7, 25=4, 32=4, 4^(-1)=2. (4^(-1) modulo 7 is, by definition, the number that, when multiplied by 4, gives product 1; hence, 4^(-1) modulo 7 is 2 modulo 7)

Putting all this together and working modulo 7:

5^(2n)+3*2^(5n-2) = 25^n + 3*4^(-1)*32^n = 4^n + 6*4^n = 4^n(1+6)=0.

This means that 5^(2n)+3*2^(5n-2) is divisible by 7.

Answer 2

I assume you want to show this for n = 1,2,....

I'm going to rewrite it by replacing n with n+1, and then showing it's true for n=0,1,2,...

You get 5^(2n+2) + 3*2^(5n+3)
which can be written as
25*25^n + 24*32^n.

Use the fact that a^n is congruent to (a+b)^n, mod b. Also, 25 = 4 mod 7 and 24 = 3 mod 7. So, if 25^n = 32^n = r mod 7, it follows that

25*25^n + 24*32^n = (3*7+4)(7p+r)+(3*7+3)(7k+r). The only term that does not explicitly include 7 is 4r+3r=0 (mod 7).

Probably more detail is needed but it's a blueprint.

Answer 3

I bet you spend a lot of time by yourself