Prove that there are infinitely many primes $p$ such that $x^{10} + x + 1 \equiv 0 \mod p$ has at least one solution $x\in\mathbb{Z}$.

Question

Prove that there are infinitely many primes $p$ such that

$$x^{10} + x + 1 \equiv 0 \mod p$$

has at least one solution $x\in\mathbb{Z}$.

I believe I should be doing a proof by contradiction but I cannot figure out where it arises. Any help will be appreciated! thank you!

Robert Z · Accepted Answer · 2018-08-30T16:04:36.703

9

Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,\dots, p_n$ then let $x:=p_1p_2\cdots p_n-1$. It follows that for $i=1,\dots,n$, $$N:=x^{10}+x+1\equiv 1-1+1=1 \pmod{p_i}.$$ Now consider $q$ be a prime which divides $N$. What may we conclude?

edited Aug 30 '18 at 16:04

answered Aug 30 '18 at 07:49

Robert Z

145,942

We can also take any finite set $S$ of primes such that $3\in S$ and let $x=\prod_{p\in S}p.$.............+1 – DanielWainfleet Aug 30 '18 at 15:31

Prove that there are infinitely many primes $p$ such that $x^{10} + x + 1 \equiv 0 \mod p$ has at least one solution $x\in\mathbb{Z}$.

1 Answers1

Linked