# Need help

$\gcd(a^{2^{m}}+1,a^{2^{n}}+1 )=\begin{cases} 1 \text{ if }a \text { is even} \\ 2 \text{ if }a \text { is odd} \end{cases}$

If $$m \neq n$$, prove the equation above.

Note by Akshat Sharda
2 years, 3 months ago

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Note that $$a^{2^{m+1}} - 1 = (a^{2^m} - 1)(a^{2^m} + 1)$$. Thus, if $$n > m$$, then $$a^{2^n} - 1 = (a^{2^m} - 1)(a^{2^m} + 1)(a^{2^{m+1}} + 1) \ldots (a^{2^{n-1}} + 1)$$. Thus $$a^{2^n} + 1 = (a^{2^m} + 1) \cdot k + 2$$ for some integer $$k$$. We know that $$\gcd(a, b+ka) = \gcd(a, b)$$, thus $$\gcd(a^{2^m} + 1, a^{2^n} + 1) = \gcd(a^{2^m} + 1, 2)$$. If $$a$$ is odd, clearly $$a^{2^m} + 1$$ is even, so the GCD is 2. If $$a$$ is even, clearly $$a^{2^m} + 1$$ is odd, so 2 doesn't divide it, thus the only remaining choice is that the GCD is 1.

- 2 years, 3 months ago

Aah! Thanks!

- 2 years, 3 months ago

Was this your original ? Nice one @Ivan Koswara !!!

- 2 years, 3 months ago

No :-P if it would have been why would I write need help

- 2 years, 3 months ago

to prove it :P

- 2 years, 3 months ago

I had a new Idea !!!

Let us research some properties of gcd of two towers leaving same remainder modulo some number

- 2 years, 3 months ago

I can't understand what you're saying! :-) But let's discuss... Are you making another note? @Chinmay Sangawadekar

- 2 years, 3 months ago

wait I will create a note , can you come on slack now ?

- 2 years, 3 months ago