Question involving Fermat's Little Theorem

Question

Q: Show that if $p$ is an odd prime, then $2p$ divides $(2^{2p-1}-2)$.

Here is how I approached the problem. Let $m=2p$. Then, $$2^{m-1}-2\pmod m$$ By Fermat's Little Theorem, $$2^{m-1}-2\equiv 1-2\equiv -1\pmod m$$

For $m$ to divide $(2^{2p-1}-2)$ I would expect the congruence to be $0$. From which I would conclude that $2p$ does not divide $(2^{2p-1}-2)$. What am I doing wrong?

Fermat’s little theorem applies to primes; for composites, use Euler’s theorem — J. W. Tanner, Jul 07 '24 at 22:48
By Fermat, $p$ prime $\Rightarrow\bmod p!:\ 2^{2p-1} = 2(2^{p-1})^2\equiv 2(1)^2\equiv 2.,$ Equivalently apply mod order reduction in the dupe, i.e. $,\color{#c00}{2p-1\equiv 1}\pmod{p-1}\Rightarrow 2^{\color{#c00}{2p-1}}\equiv 2^{\color{#c00}1}\pmod{p}.\ $ Your error is attempting to apply little Fermat for non-prime modulus. — Bill Dubuque, Jul 07 '24 at 23:46
Please delete the question once all is clear, since we already have too many questions like this. — Bill Dubuque, Jul 07 '24 at 23:47
Thank you, @BillDubuque. I searched for applicable questions but didn't search the correct terms. Will do as you say. — k endres, Jul 08 '24 at 01:47

score 1 · Answer 1 · answered Jul 07 '24 at 22:54

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As mentioned in the comment, Fermat's little only applies to primes.

Since $p$ and $2$ are coprime, it suffices to show $2\mid 2^{2p-1}-2$ (which is obvious as both of $2^{2p-1}$ and $2$ are even), and $p\mid 2^{2p-1}-2$, for which we may apply Fermat's little:

$$2^{p-1}\equiv 1\Rightarrow 2^{2p-1}=2^{2(p-1)+1}\equiv (2^{p-1})^2\cdot 2\equiv 2$$

answered Jul 07 '24 at 22:54

Just a user

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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jul 07 '24 at 23:51

score 1 · Answer 2 · answered Jul 07 '24 at 22:55

1

What you are doing wrong is applying Fermat’s Little Theorem to $m=2p$, which is not prime. For composites, you should use Euler’s theorem.

answered Jul 07 '24 at 22:55

J. W. Tanner

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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jul 07 '24 at 23:51
The proposed duplicate question is different since it asks about prime moduli, and while one of the answers there mentions Euler, it doesn’t answer the question here (viz., what OP here did wrong was applying Fermat’s little theorem to a composite modulus) – J. W. Tanner Jul 08 '24 at 01:30
But we only need the prime moduli case to deduce the result - see my comment on the question. This site is not meant for collecting mistakes (arithmetic errors, theorems wrongly invoked by forgetting hypotheses, etc). Questions based on such oversights are not of general interest and only serve to clutter search results so they should be deleted. Answering them makes it much more work to delete them. Better to point out the oversight in a comment - not an answer. – Bill Dubuque Jul 08 '24 at 01:43
Let's list all axioms, and call everything else "dupe". – Just a user Jul 08 '24 at 04:19

Question involving Fermat's Little Theorem

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