One more sequence with infinite prime divisors

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One more sequence with infinite prime divisors

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#1 h Apr 27, 2025, 1:01 PM

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Prove that there are infinitely many primes such that each divides an integer of the form $2^n + 3^{n+2} + 5^{n+1}$ for some positive integer $n$ .

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#2 h Apr 27, 2025, 3:06 PM • 1 Y

Y by tom-nowy

Assassino9931 wrote:

Prove that there are infinitely many primes such that each divides an integer of the form $2^n + 3^{n+2} + 5^{n+1}$ for some positive integer $n$ .

The key equality is $2-27+25=0$ .
Suppose that $p$ is a prime number such that $\left( \frac{2}{p} \right)=\left( \frac{5}{p} \right)=1$ and $\left( \frac{3}{p} \right)=-1$ . Let's take $n=\frac{p-1}{2}+1$ . Then $2^n \equiv 2, \quad 3^{n+2} \equiv -27, \quad 5^{n+1} \equiv 25 \pmod p$ , because $a^{\frac{p-1}{2}} \equiv \left( \frac{a}{p} \right) \pmod p$ . So $2^n + 3^{n+2} + 5^{n+1} \equiv 2-27+25=0 \pmod p$ .

So it's enough to check if there are infinitely many such primes $p$ . Take $p=120t+41$ (there are infinitely many such primes by Dirichlet's theorem). By QLR we have $(2/p)=(5/p)=1$ and $(3/p)=(p/3)=-1$ .

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