I suspect the negation of this statement is true. I was looking at this problem for fixed $\alpha,\beta$ but that's really hard, so I wanted to prove that there is at least one pair of integers with the property. So, supposing the opposite of that: for all $\alpha,\beta\in\mathbb{N}$ the set of primes satisfying the property $\gcd(\text{ord}_p(\alpha),\text{ord}_p(\beta))=1$ is bounded. So we can define a function $f(n)$ to be the greatest prime with the desired property over all pairs of integers $\alpha,\beta\leq n$.

Increasing $n$ appends more pairs of integers to the numbers over which $f$ has to satisfy its definition, plus the ones that were $\leq n$, so $f$ is nondecreasing. Also for a prime $p\equiv 3 \mod 4$ we can take $\alpha\equiv -1$ and $\beta\equiv w^2$ where $w$ is a primitive root and we have $\gcd(\text{ord}_p(\alpha),\text{ord}_p(\beta))=\gcd\left(2,\frac{p-1}2\right)=1$ so $f$ is unbounded.

So for some large $n$ and any prime $p>f(n)$ we have that of the $(p-1)^2$ pairs of numbers in $(\mathbb{Z}/p\mathbb{Z})^\times$ we have a forbidden square in the bottom left $(n-1)^2$ elements whose elements have to have pairwise noncoprime orders. If $p-1$ has $k$ distinct prime divisors, there are $2^{k-1}$ ways of writing it as $p-1=\theta\phi$ with $\gcd(\theta,\phi)=1$ and with a primitive root $w$ we can generate $w^\theta$ and $w^\phi$ with our property. So for each of those exponentially increasing quantity of pairs of numbers, they have to all conspire to be outside the forbidden square for all primes greater than $f(n)$. This just seems unsustainable. $k$ isn't always large, for example if $p$ is a Sophie Germain prime $k=2$ and for all we know there are infinitely many of those, however it is large often enough to be a problem.

Can we derive a contradiction from this?

Some computational data:

For $(\alpha,\beta)=(2,3)$ the primes with the desired property are 683, 599479, 108390409, 149817457, 666591179, 2000634731, 4562284561, 14764460089, 24040333283 and none more up to $1752\cdot10^9$.

For $(2,5)$ the primes are 31, 601, 2593, 599479, 204700049, 466344409, 668731841, 11638603429.

For $(3,5)$, they are 13, 313, 51169, 797161, 3482851, 5096867, 12207031, 162410641, 368385827, 1001523179, 4902814883.

And for $(2,6)$, they are 5, 7, 31, 43, 135607, 153649, 270841, 1489441, 1505447, 25781083, 127236649, 558062249, 745988807, 27989941729, 29512739491, 47206579351.

  • $\begingroup$ Isn't $ord_p(\alpha) = ord_p(\beta) = 0$ for all but finitely many primes $p$? $\endgroup$
    – D_S
    Jul 4, 2021 at 16:41
  • $\begingroup$ @D_S Here $\text{ord}$ doesn't refer to the p-adic order. It is meant to represent the order of an element in the multiplicative group $(\Bbb{Z}/p\Bbb{Z})^{\times}$. $\endgroup$
    – arbashn
    Jul 4, 2021 at 16:54


You must log in to answer this question.

Browse other questions tagged .