S.O.S. Mathematics CyberBoard

I was wondering if you could help me solve an induction problem... I have to prove that for any non negative integer n, the following is true:

7^(n+2)+8^(2n+1) = 0(mod 57)

*note "=" sign represents congruent to

First start off by considering the case n = 1. I think you can verify it using a calculator.

Then, assume the formula is true for n = k:

7^(k+2) + 8^(2k+1) = 0(mod 57)

Now, try to prove it for n = k+1

7^(k+3) + 8^(2k+3) = 7*(7^(k+2)) + 64*(8^(2k+1))

If you can somehow isolate 7^(k+2) + 8^(2k+1), it should be easy from there.

Can anyone show us how to issolate them?!?!?!?!?!?!?!?!?!?

You have already asked this question here.

True, but now he's trying to prove it by induction, so he will need a different method. Unfortunately, I probably don't know enough about modular math to isolate
7^(k+2) + 8^(2k+1) and hence solve the problem.

nada, see this



and ask yourself what is 64 mod 57.

I am also trying to work on this problem and I do not understand how the fact that 64 = 7 mod 57 helps to conclude that the whole thing is congruent to 0 mod 57

7*(7^(k+2)) + 64*(8^(2k+1)) = 7*(7^(k+2)) + (7 + 57)*(8^(2k+1)) = 7*(7^(k+2) + 8^(2k+1)) + 57*(8^(2k+1))
Can you finish now?

Yes, thanks!