Let $a$ be an odd integer. Prove that $a^{2^m}+2^{2^m}$ and $a^{2^n}+2^{2^n}$ are relatively prime for all positive integers $n$ and $m$ with $n\not= m$.
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Tags: modular arithmetic, number theory, relatively prime
pluristiq
09.01.2010 04:25
it's wrong: for $ (a,m,n) = (3,3,7)$, both expressions are divisible by 5. what you meant to say is $ a^{2^m} + 2^{2^m}$ and $ a^{2^n} + 2^{2^n}$. we may assume WLOG that $ m < n$. let $ p|a^{2^m} + 2^{2^m}$, where $ p$ is prime. since $ a$ is odd, $ p$ is odd. then $ a^{2^m}\equiv - 2^{2^m} \pmod{p}$. repeatedly squaring both sides, we eventually get $ a^{2^n}\equiv 2^{2^n}\not\equiv - 2^{2^n}\pmod{p}$, since $ p$ is odd. this shows that the sets of primes dividing the two expressions are disjoint, implying the assertion.
HadjBrahim-Abderrahim
17.08.2016 18:58
hasan4444 wrote: Let $a$ be an odd integer. Prove that $a^{2^m}+2^{2^m}$ and $a^{2^n}+2^{2^n}$ are relatively prime for all positive integers $n$ and $m$ with $n\not= m$.
Without loss of generality, assume that $n> m.$ it is easy to see that, $a^{2^m}\equiv -2^{2^m} \pmod{p}$, for every prime $p$ dividing $a^{2^m}+2^{2^m}.$ So we deduce that,
$$a^{2^{n}} \equiv \left(a^{2^m} \right)^{2^{n-m}} \equiv \left(-2^{2^m} \right)^{2^{n-m}} \equiv 2^{2^n} \pmod{p}$$This mean that, $p \mid a^{2^n}-2^{2^n}$ for every positve integer $n >m.$ Now, Assume that $\gcd \left(a^{2^m}+2^{2^m},a^{2^n}+2^{2^n} \right) \neq 1$, For some positive integer $n >m.$ This, implied that there exist a prime divisor of $a^{2^m}+2^{2^m}$ devide $a^{2^n}+2^{2^n}$ so this divisor must devide $\left(a^{2^n}+2^{2^n}\right)-\left(a^{2^n}-2^{2^n} \right)=2^{2^{n}+1}.$ Since, $a$ is an odd integer we deduce that, $a^{2^m}+2^{2^m}$ is an odd positive integer. Thus, $p \neq 2$ which mean that $p \nmid 2^{2^n+1}$ for any positive integer $n.$ So we deduce the contradiction. Therefore, by symmetric we deduce that, $\gcd \left(a^{2^m}+2^{2^m},a^{2^n}+2^{2^n} \right)=1$ for all positive integers $n$ and $m$ with $n\not= m.$