Modular ??

Question

Let m be an odd integer. Show m^(2^n) ≡1 mod 2^(n+2) for all positive natural numbers n.

So, I'm pretty sure that m^(2^n) is odd, while 2^(n+2) is even. But then I'm not sure how to go about showing this...

Answer 1

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Claim: If m is an odd number, then m^n is odd for all natural numbers n.

1) Let n = 1. Then m^1 = m, and m is obviously odd.
2) Assume that the formula holds true for up to n = k, for some value k. Then by our induction hypothesis, m^k is odd.

{We want to prove that m^(k + 1) is odd}.

But, what is m^(k + 1) equal to?
It's equal to (m^k) * (m^1), or
(m^k) * m

We know that m^k is odd by our induction hypothesis, and we know that m is odd. The product of two odd numbers is an odd number.

Therefore, (m^k) * m is odd, which is equal to m^(k + 1).
This proves that m^(k + 1) is odd.

Therefore, by the principal of mathematical induction,
m^n is odd for all natural numbers n.
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If m^n is odd for all natural numbers n, then m to *ANY* natural number is odd.

Since 2^n is a natural number, it follows that
m^(2^n) is odd.

We also know that 2^(n + 2) = 2 * 2^(n + 1), which shows that
2^(n + 2) is even.

I'm not sure where to go from here. But you are correct that
m^(2^n) is odd.

Answer 2

You have the basic idea.


I'd suggest appling the definition of modulo:

Let m = 2k+1 and raise it to some powers of 2^n and fiddle with it!

Answer 3

Represent the coefficient of the variable as being equivalent to the cube root of log (y/x + 4.57).