Equivalence of multiplicative order modulo prime

Asked Sep 17 '24 at 13:48

Active Sep 17 '24 at 14:36

Viewed 78 times

Given $2$ positive integers $a, b$. If for each prime $p$ such that $p\nmid a$ and $p\nmid b$, the multiplicative order of $a$ modulo $p$ always equal the multiplicative order of $b$ modulo $p$, does it always have $a=b$?

It is straightforward to see that if $a, b$ satisfy the condition, the set of prime divisors of $a^n-1$ always coincides with the set of prime divisors of $b^n-1$ for all $n\in\mathbb{Z}^{+}$. However, I currently have no idea how to proceed further.

edited Sep 17 '24 at 14:36

asked Sep 17 '24 at 13:48

Wembley Inter

What does the notation $[a, b]$ mean in this context? – Tzimmo Sep 17 '24 at 13:58
@Tzimmo In the context of Number Theory, the notation $[a,b]$ is often used for the least common multiple of $a$ and $b$ – jjagmath Sep 17 '24 at 14:03
You probably meant $\color{red}{b}^n-1$ the second time. – jjagmath Sep 17 '24 at 14:06
Ah, I've only seen the round parentheses for gcd. Thanks. – Tzimmo Sep 17 '24 at 14:20
@jjagmath Thanks for pointing that out, already edited. – Wembley Inter Sep 17 '24 at 14:36
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Wow, intuitively the cases $n = 1$ and $n = 2$ together should be enough to show that $a = b$ let alone when you have all $n$ but indeed it is not obvious at all how to show this. Very interesting question! I will think about it – Vincent Sep 17 '24 at 14:41
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Vincent, p = 5 and q = 17 are two primes such that p-1 and q-1 have the same prime factors and p^2 - 1 and q^2 - 1 have the same prime factors. However, on the computer I have not found one a pair (p,q) like that where I include also the condition n = 3. – tkr Sep 17 '24 at 14:51
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While I'm hopeful there's an unconditional proof, I'll point out that the $abc$ conjecture does imply $a=b$: if not, WLOG $a<b$, and then we have $(b^n-1)+1=b^n$ where the radical of $b^n-1$ is at most $a^n-1$, and thus the product of the three radicals is at most $a^nb = (b^n)^{(\log a)/\log b+1/n}$, and the exponent approaches $(\log a)/\log b < 1$ as $n\to\infty$. – Greg Martin Sep 17 '24 at 16:23
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This question is similar to: Is an integer uniquely determined by its multiplicative order mod every prime. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. Found using an Approach0 search. ... – John Omielan Sep 17 '24 at 16:34
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FYI, I found this question through links in the AoPS threads I believe it's okay, but I can not solve and ord(a)=ord(b) for any prime p. Also, the AoPS thread Equal orders asks a related question. – John Omielan Sep 17 '24 at 16:43
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@JohnOmielan Thanks for your notification! I've learned to use Approach0 next time. – Wembley Inter Sep 17 '24 at 17:12

Equivalence of multiplicative order modulo prime

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