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

If \(m \neq n\), prove the equation above.

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## Comments

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TopNewestNote 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.

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Aah! Thanks!

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

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I had a new Idea !!!

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

@Akshat Sharda

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

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

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