3

I need to show that, for each $n \geq 2$, the whole number $4n^{2}-1$ is a Fermat pseudoprime to base $2n$; (hint: $4n^{2}-2$ is a multiple of $2$).

Fermat pseudoprime to base $2n$ means that $(2n)^{4n^{2}-2} \equiv 1 \mod 4n^{2}-1$, but from this point on I don't know what to do.

Thanks.

DrMontorsi
  • 31

2 Answers2

3

Another hint besides the one given:

$$(2n)^2\equiv 1\bmod 4n^2-1$$

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
1

$a$ is a Fermat psp base $m\iff\bmod m\!:\ a^{m-1}\equiv 1\iff {\rm ord}(a)\mid m\!-\!1,\,$ by the Order Theorem

Thus in OP we need: $\ {\rm ord}(2n)\mid 4n^{\large 2}-2 = \color{#c00}2(2n^{\large 2}-1)\,$ with an obvious $\rm\color{#c00}{small\ factor}$ to try first.

It works: $\!\bmod 4n^{\large 2}\!-1\!:\,\ 4n^{\large 2}\equiv 1\,\Rightarrow\, (2n)^{\large\color{#c00}2}\equiv\ \ldots $

Bill Dubuque
  • 282,220
  • 1
    This yields a conceptual way to discover it. Another conceptual way is to note that $,4n^2-1 = (2n-1)(2n+1)$ and mod each factor $2n\equiv \pm1$ so has order at most $\color{#c00}2,,$ so ditto mod their product, by CRT – Bill Dubuque Dec 26 '19 at 22:55