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Solve in $\mathbb N$: $ 2^x\equiv 1 \bmod{x}$.

1) first we showed that for all $ m $ and $ n \in\mathbb N$ if $\gcd(m,n)=d$ then ($2^m\equiv 1 \bmod{p} $ and $2^n\equiv 1 \bmod{p} $) implies that $2^d\equiv 1 \bmod{p}$ for all $p$ $\in\mathbb N^*$.

2) if we consider that $p$ is the smallest prime number dividing $x$ we would have $\gcd(x-1,p) = 1$ and $2^{p-1} \equiv 1 \bmod{p} $ (using Fermat) and $ 2^x \equiv 1 \bmod{p}$.

Can I deduce something from this ?

user26857
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LeFo
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2 Answers2

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I did not look in too many details but : First note that $x=0$ and $x=1$ are trivial solutions.

Then note that for the equations to hold with $x>1$ you need $x$ to be odd, because $2^x$ is even.

Finally using Fermat, we know that for any odd prime $p$, we will never have the equality as $$2^p\equiv2 \mod p$$

Therefore using your proposition, any two numbers verifying the equations would need to be relatively prime (otherwise they would have at least common denominator prime, your proposition would induce $2^p\equiv 1 \mod p$, which is false by Fermat). But then for that to hold for ANY two number, they must be prime themselves. A contradiction from Fermat again.

So I'm guessing your problem has no solutions except the trivial ones. A rigourous proof would require some substantial addition though.

Thomas Lesgourgues
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  • you can't say that $0$ is a solution because $\mod{0} $ doesn't exist – LeFo Feb 06 '19 at 11:30
  • That's depend on your convention. If you define $a \equiv b \mod p \Leftrightarrow a=b+kp$, then modulo 0 can exist, and is equivalent to $a=b$. There is a full discussion here https://math.stackexchange.com/questions/516251/why-is-n-mod-0-undefined

    In that sense we do have $2^0\equiv 1 \mod 0$

     – Thomas Lesgourgues Feb 06 '19 at 11:35
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For $x=p$, we have by Fermat's little theorem that $2^p\cong 2\not\cong1\pmod p$, so $x$ is not a solution.

So $x$ can't be prime.