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Let $p_1,p_2$ be primes and $x\in\mathbb{N}$. I want to investigate \begin{equation*} p_1^x\equiv p_2^x\equiv 1 \pmod{p_1p_2-1} \end{equation*} I want to find how $x$ depends on $p_1$ and $p_2$.

This is clearly something to do with the order of the primes but I can't find anything on comparing primes with the same order.

One way I thought to do this was to look at it as \begin{equation*} p_1^x-1\equiv p_2^x-1\equiv 0\pmod{p_1p_2-1} \end{equation*} and look at how their cyclotomic polynomials interact with each other, but cyclotomic polynomials don't really have any results for this.

For specific values this is easily evaluated. For example; $p_1=7$ and $p_2=11$, we get

\begin{equation*} 7^6=117649\equiv11^6\equiv1771561\equiv1\pmod{76} \end{equation*} so here $x=6$.

Any advice would be very appreciated.

Arturo Magidin
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MeBadMaths
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    The equation holds for any two primes $p_1$ and $p_2$ with $x$ a multiple of $\varphi(p_1 p_2 - 1)$ where $\varphi$ is Euler's totient function. – Oussema Apr 30 '21 at 14:37
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    This is a direct result of Euler's theorem because $p_1$ and $p_1 p_2 - 1$ are always coprime. – Oussema Apr 30 '21 at 14:39
  • In fact, you can generalise this fact to any two integers $m$ and $n$. – Oussema Apr 30 '21 at 14:40
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    You can do better by taking $x$ to be a multiple of $\lambda(mn -1)$ where $\lambda$ is Charmichael's lambda function. – Oussema Apr 30 '21 at 14:42
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    The smallest such $x$ is $\operatorname{lcm}(\operatorname{ord}{mn-1}(n), \operatorname{ord}{mn-1}(m))$ where $\operatorname{ord}_{a}(b)$ is the order of $b$ modulo $a$. That is the smallest integer $t$ such that $b^t \equiv 1 \pmod{a}$. – Oussema Apr 30 '21 at 14:44
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    Use \pmod{p_1p_2-1} to get the parenthetical mod operator. – Arturo Magidin Apr 30 '21 at 14:47
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    Thank you @Oussema for your responses. I hadn't noticed that $p_1$ and $p_1p_2-1$ were coprime, that's great! I will look into Charmichael's Lambda function.

    I am mainly looking at primes, though you're right this can be generalised to any intergers $m$ and $n$.

     – MeBadMaths Apr 30 '21 at 14:55
  • And thank you @ArturoMagidin I didn't know that piece of code - that will absolutely help for my write up too... – MeBadMaths Apr 30 '21 at 14:55

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$\!\!\bmod p_1 p_2\! -\!1\!:\,\ p_1 p_2\equiv 1\Rightarrow p_2\equiv p_1^{-1}\,$ so $\,p_1^n\!\equiv\! 1\iff p_2^n\equiv 1,\,$ so your question boils down to finding the order of $p_1$ (= order of $p_2),\,$ necessarily a divisor of every $n$ such that $\,p_1^n\equiv 1\,$ (e.g. the order must divide the Euler phi and Carmichael lambda function of the modulus).

Bill Dubuque
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