Studying for an upcoming comprehensive exam we stumbled upon the following problem:

Prove that for every positive integer $a$, there exists infinitely many composite integers $n$ such that $a^{n-1}\equiv 1 \mod n$. (Hint: Choose $n=\frac{a^{2p}-1}{a^2-1}$ for a suitable prime $p$.)

We have tried many approaches to no avail. The most promising approach involved choosing $p$ such that $a$ is a quadratic residue modulo $p$. Then $n-1=\frac{a^2(a^{2(p-1)}-1)}{a^2-1}$ has a factor of $p$ (the second factor in the numerator is a difference of squares which factors to another difference of squares and finally an $a^{(p-1)/2}-1$ pops out which by our choice of $p$ must be divisible by $p$. We were unable to conclude from here. Any help would be appreciated.

  • $\begingroup$ Won't this always be the case whenever $\gcd(a, n) = 1$? $\endgroup$ Aug 14 '19 at 21:27
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    $\begingroup$ @RobertShore if $\gcd(a, n) = 1$ we get $a^{\varphi(n)} \equiv 1 \bmod n$ but $\varphi(n)$ may not be $n - 1$ or even divide it. $\endgroup$ Aug 14 '19 at 21:36
  • $\begingroup$ @RobertShore $3^{4-1}=27 \equiv 3 \pmod{4}$$ $\endgroup$
    – N. S.
    Aug 14 '19 at 21:37

First: If $p$ is a prime and $$n=\frac{a^{2p}-1}{a^2-1} =1+a^2+a^4+...+a^{2p-2}$$

Then $$ 1+a^2+a^4+...+a^{2p-2}=0 \pmod{n} \\ a^2( 1+a^2+a^4+...+a^{2p-2})=0 \pmod{n} \\ a^2+a^4+...+a^{2p-2}+a^{2p}=0 \pmod{n} \\ $$

Subtracting the first and last relation you get $$1=a^{2p} \pmod{n}$$

Now, if you now chose $p,n$ so that $2p |n-1$ you get $$a^{n-1} \equiv 1 \pmod{n}$$

Note that $2p |n-1$ means $$2p|a^2(1+a^2+...+a^{2p-4})$$

It is easy to make the RHS even and if you want $$p|1+a^2+...+a^{2p-4}$$ the easiest way is to pick $p >a^2-1$ (to make sure that $a^2-1$ cannot be divisible by $p$) and use $$a^{2p-2} \equiv (a^{p-2})^2\equiv 1 \pmod{p}$$

  • $\begingroup$ Wow! This is excellent. Thank you! $\endgroup$
    – Drew Meier
    Aug 14 '19 at 22:05

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