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If $m > n$ and $a,m,n$ are positive, with $m$ not equal to $n$, find the greatest common divisor of $2^{2^m}+1, 2^{2^n}+1$. Please solve this problem using Euclid's algorithm.

I tried to use Euclid's algorithm by first dividing $2^{2^m}+1$ by $2^{2^n}+1$, but I got stuck in the 2nd step while dividing $2^{2^n}+1$ by the difference of $2^{2^m}+1$ and $2^{2^n}+1$.

skallu
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    https://math.stackexchange.com/questions/123524/fermat-numbers-are-coprime – lab bhattacharjee May 15 '19 at 05:36
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    There is no such thing as a "greatest common denominator". Two fractions will have a least common denominator. Two integers will have a greatest common divisor. Anyway, do you know what Euclid's Algorithm is? Do you know how to apply it? Where do you get stuck when you try to apply it to this question? We can't help you, if we don't know what you can do, and what you can't do. – Gerry Myerson May 15 '19 at 06:13
  • As written, the answer is $(a^2)^{n+1}$, obviously. PLease check the statement – Hagen von Eitzen May 15 '19 at 06:13
  • @Hagen, look at the edit history. A comedy of errors. – Gerry Myerson May 16 '19 at 03:19
  • @Garry and Hagen, sorry for the confusion that I have caused. Being new to this website, I have many things to learn. I have rightly edited the problem now and if you can, please help me. – skallu May 16 '19 at 04:25
  • Have you not noticed, Rohan, that your question has been closed? That means no one can post an answer. But, luckily for you, it has been closed as a duplicate, and a link has been furnished to the earlier question (in fact, to three earlier questions) that your question duplicates. So have a look at those three earlier questions, and see whether they settle things for you. – Gerry Myerson May 16 '19 at 23:33
  • @Garry Sorry again :( – skallu May 17 '19 at 00:44
  • So, have you had a look at those earlier questions, Rohan? Are things settled now? – Gerry Myerson May 18 '19 at 05:08
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    @Garry Thanks for asking. I am now clear on this topic. I am now trying to further improve my understanding on Euclid's algorithm by solving more problems like this. – skallu May 18 '19 at 11:43

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The first step in Euclid's algorithm:

If we let $a_k = 2^{2^k}+1$, and

$$g(m,n)=\sum _{j=1}^{2^{m-n}} (-1)^{j+1}2^{2^m-j 2^n}$$

then

$$a_m=a_ng(m,n)+2$$

The second step would be

$$a_n=2\left(\frac{a_n-1}{2}\right)+1$$

which shows that $a_m,a_n$ are coprime for $m>n>0$

Lozenges
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