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Prove that two distinct numbers of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if $a$ is even and have $\gcd=2$ if $a$ is odd.

My attempt:
If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$
Then, $$a^{2^{n}} + 1 = 2^{2^{n}s}\cdot k^{2^n} + 1$$ and $$a^{2^{m}} + 1 = 2^{2^{m}s}\cdot k^{2^m} + 1$$ To prove that they're relatively prime, we need to show that their gcd = 1. And I was stuck here, how could I prove that gcd of two numbers is $1$?

A hint would be sufficient. Thanks.

user26857
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roxrook
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3 Answers3

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If $x = -1 \mod p$, then $x^{2^n} = 1 \mod p$.

Assume $n \gt m$. So if $p$ divides $x + 1 = a^{2^m} + 1$, then, $a^{2^n} + 1 = x^{2^{n-m}} + 1 = 1 + 1 = 2 \mod p$.

Aryabhata
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  • @Moron: What's $p$ in your hint? I tried understand your hint, but I felt lost :(. Thanks. – roxrook Feb 28 '11 at 23:40
  • @Chan: $p$ is any prime which divides $x+1$. Basically we are showing that if $p$ divides $a^{2^m} + 1$, then $p$ cannot possibly also divide $a^{2^n} + 1$, unless $p=2$. – Aryabhata Mar 01 '11 at 00:09
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    @Moron: Thanks again. But I lost, I knew that $p$ is prime, but I can't find a way to link it to my problem. Furthermore, how does it relates to the parity of $a$? – roxrook Mar 01 '11 at 04:55
  • @Chan: If two number are not relatively prime, there is some prime $p$ which divides both. We showed that that is not possible for $p \gt 2$. If a is odd, then all numbers are divisible by 2. – Aryabhata Mar 01 '11 at 05:40
  • @Moron: Thank you. I will think more about it. – roxrook Mar 01 '11 at 05:51
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Hint:

Consider the following proof when $a=2$: https://planetmath.org/fermatnumbersarecoprime

Try adapting it to work for all $a$.

Hint 2: Factor $a^{2^n}-1$.

user26857
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Eric Naslund
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    @Eric Naslund: I tried to adapt the proof you mentioned, but I could not prove that the gcd must be 2, since $a$ has another variables :(. Could you help? – roxrook Feb 28 '11 at 23:50
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    @Chan: Ok, so we need to do a little more, you are correct. It suffices to consider $$a^{2^n}+1$$ modulo $4$. Since $$a^{2^n}$$ is a square, it is congruent to 0 or 1, and hence is congruent to 1 when $a$ is odd. Then we see $$a^{2^n}+1\equiv 1\pmod{4}$$ so that each term is divisible by 2 but not 4. That is where the $2$ in the $\gcd$ comes from. Use the previous method to show that other primes cannot divide it. – Eric Naslund Mar 01 '11 at 06:17
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    @EricNaslund,@Chan: What about the case of a is even? – Alexy Vincenzo Jun 29 '13 at 19:11
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Proof $\ (A\!+\!1, A^{2K}\!\!+1)\,=\, (A\!+\!1,\ \overbrace{(-1)^{2K}\!+1}^{\large 2})\ \ $ [OP is $\, A = a^{\large 2^{\Large N}}\!\!,\,$ wlog $N\! <\! M$]

using $\,\ \ (A\!+\!n,\, \ F(A)\:\!\ )\, = \, (A\!+\!n,\ F(-n) )\ $ for all polynomials $ F(X)\in \mathbb Z[X],\ A,n\in \mathbb Z$

Remark $ $ For general case $\ (a^m+1,a^n+1)\ $ see this answer.

Bill Dubuque
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    Someone is downvoting all the answers on this thread. +1. This is a more generally applicable answer. – Aryabhata Feb 28 '11 at 05:18
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    At this point, I think you under evaluate a very important part of mathematics. That is the ability to communicate it to other people. Every answer you down voted was one where I was much clearer to someone wanting to understand the basics. Each answer you write isn't even written in English nicely, is unclear, and not at the level the OP wanted. Having said that, do I think they are bad answers that deserve to be down voted? No, that I reserve for actual things that do not contribute at all mathematically. – Eric Naslund Feb 28 '11 at 16:20
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    At the very least, if you do not like my answers, and think they need to be pushed in a certain direction, please comment and tell me. Whenever you down vote a message box comes up asking you to do so. It is impossible for me understand your reasoning when you do not tell me. I do not see at all what was wrong with most of these answers, and I honestly have no problem with downvotes that are explained, because then whatever problem there is can be improved. As is it resembles that you just do not like the fact that mine are accepted over yours, which is childish. – Eric Naslund Feb 28 '11 at 16:28
  • @readers Eric's answer now consists of a dead link. Speculation about downvotes is usually wrong, almost always nonconstructive, and always off-topic. Worth emphasis: this answer is the only answer with a complete, rigorous proof (and even a generalization) - something we should strive for in answers. Yet someone downvoted it (presumably explained by the above strange comments). Alas, gamification is the root of all evil. – Bill Dubuque Apr 28 '20 at 15:55
  • Hello @Bill Dubuque , not a downvoter ( I don't even have that ability yet), but reading your answers, not this one in particular, but quite a few, if I already know the theorem you are talking about, they are almost always the most elegant, generalised and beautiful possible answers. However, for areas that I am new towards, I do feel your answers are slightly unhelpful, and scary to see, often requiring me to search through many additional links to comprehend them, which is not always great! It would be nice if you provided a simple solution along with a generalisation. – Illusioner_ Oct 25 '24 at 11:15
  • @Illusioner More general answers usually do require more work to comprehend, but they are worth the extra work if they yield deeper insights. One should not discourage such expositions simply because they may not be (easily) comprehended by some users. The purpose of this site is not to help a single user but rather to collect a library of high-quality mathematical Q&A's that are helpful to many readers. Generally we cannot know the knowledge level of such readers. – Bill Dubuque Oct 25 '24 at 20:37
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    @Illusioner Some of the most helpful texts I read as a student were those that were slightly beyond my knowledge level horizon - which planted the seeds of ideas that sparked me to learn more about the innate beauty of more advanced viewpoints. Many of my posts strive to similarly help others discover these beautiful viewpoints. – Bill Dubuque Oct 25 '24 at 20:41