# Size constraints on reduced solutions to $x^p\equiv 1\mod q$

Are there any size constraints on the reduced solutions $$x$$, $$(0, to $$x^p\equiv 1\mod q$$, for $$p$$ and $$q$$ specific primes? (Considering the primitive $$p$$-th roots of unity modulo $$q$$).

I can see that $$x>q^{1/p}$$ as $$x^p>q$$. Also for $$p=2$$, the solutions are simply $$\pm 1$$. Besides these, can anything be said so far about the lower or upper bounds?

• If $p$ is coprime with $q-1$ you can get a very good one. Jul 22, 2021 at 2:51
• That is true, I guess I was considering mainly cases where p can divide the order of elements. Jul 22, 2021 at 3:31

A very partial answer. As has been pointed out, for $$p\nmid q-1$$ the only such $$x$$ is $$1$$, and the problem is not very interesting. So, assume $$p\mid q-1$$.
Let $$1 be a non-unit solution to $$x^p\equiv 1\pmod q$$, and define $$x_n=x_1^n\operatorname{rem}q$$ ($$\operatorname{rem}$$ denotes remainder), so that $$\{1=x_0,x_1,\dots,x_{p-1}\}$$ are the unique values of $$x$$ between $$0$$ and $$q$$ satisfying $$x^p\equiv 1\pmod q$$. We have $$1+x_1+x_2+\cdots+x_{p-1}\equiv 0\pmod q.$$ This implies that $$\max_{1\leq i\leq p-1}x_i\geq \frac{q-1}{p-1}\qquad \min_{1\leq i\leq p-1}x_i\leq q-1-\frac{q-2}{p-1}.$$ This isn't a particularly strong bound, but it's better than $$q^{1/p}$$. (Ifyou want, by using that the $$x_i$$ are distinct, you can improve it by about $$p/2$$, which does nothing for large $$q$$ and fixed $$p$$ but might be interesting if $$p$$ grows with $$q$$.)
In the special case of $$p=3$$, the values $$x_1$$ and $$x_2$$ satisfy $$x_2=p-1-x_1$$, and so we can study the distribution of all $$x_i$$ by simply studying the roots of the quadratic $$x^2+x+1$$. By a result of Duke, Friedlander, and Iwaniec, such roots have $$x_i/q$$ uniformly distributed in $$(0,1)$$ as $$q$$ grows, and so the roots can be arbitrarily closely clustered to $$p/2$$, or arbitrarily close to $$0$$ and $$1$$ (as constants times $$q$$).