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While solving the problems of chapter 1 from Introduction to Analytic Number Theory by Tom M. Apostol, I observed something interesting.

In problem 4, we show that $(a+b,a-b) = 1$ or $2$ and we have $(a+b)\cdot(a-b) = (a^2-b^2)$.

Similarly, in problem 5, we show that $(a+b,a^2-ab+b^2) = 1$ or $3$ and we have $(a+b)\cdot(a^2-ab+b^2) = (a^3+b^3)$.

I was wondering if it is true in general. My precise question is the following:

Can we expect that $(c,d) = 1$ or a factor of $n$, when $c\cdot d = a^n \pm b^n,$ where $c$ and $d$ are expression in $a$ and $b$.

And if it is true, what are those $n$?

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Bill Dubuque
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SARTHAK GUPTA
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    Consider $c=8$ and $d=10$ where $c\cdot d=3^4-1^4$ and $\gcd(c,d)=2$ – J. W. Tanner Aug 16 '20 at 20:05
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    @J.W.Tanner Thanks for pointing this out. May be I will modify the question with gcd $(c,d)$ = a factor of $n$. – SARTHAK GUPTA Aug 16 '20 at 20:08
  • It seems true at least for all primes. It boils down to proving $p\mid {p-1\choose k}+1$ when $k$ is odd and $p\mid {p-1\choose k}-1$ when $k$ is even. At least this is true for $5$ and $7$. Also if prime is true then any multiple of prime is true and the only things left is power of $2$. – cr001 Aug 16 '20 at 20:18

1 Answers1

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To continue the general examples suggested above: if $p$ is an odd prime, then $\gcd\left(a + b, \frac{a^p + b^p}{a + b}\right) \in \{ 1, p \}$.

The proof uses what is called the lifting the exponent lemma. Suppose $q$ is a prime that divides $a + b$. Then $v_q(a^p + b^p) = v_q(a + b) + v_q(p)$, i.e., $v_q\left(\frac{a^p + b^p}{a + b}\right) = v_q(p)$. If $q = p$ this quantity is $1$; otherwise it is $0$. This tells us that $q$ divides $\frac{a^p + b^p}{a + b}$ if and only if $q = p$. Thus, if $p \mid a + b$, then $\gcd\left(a + b, \frac{a^p + b^p}{a + b}\right) = p$; otherwise, $\gcd\left(a + b, \frac{a^p + b^p}{a + b}\right) = 1$.

In fact, a similar argument shows that $\gcd\left(a + b, \frac{a^n + b^n}{a + b}\right)$ divides $n$.

Paco Adajar
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  • Thanks for your answer. Can we conclude something for $n$ composite? – SARTHAK GUPTA Aug 17 '20 at 09:29
  • @SARTHAKGUPTA Well, we should be able to conclude, by a similar argument, that $\gcd\left(a+b, \frac{a^n + b^n}{a + b}\right)$ divides $n$. Maybe I should edit that into the answer. – Paco Adajar Aug 17 '20 at 17:07