Are you sure that you did not miss a condition? In particular, we would need to solve $5^y-1=4q$ i.e. to decide whether $\frac{5^y-1}{4}$ is prime infinitely often (clearly, then $y$ also has to be prime). This should be roughly as hard as the question of infinitude of Mersenne primes, in particular the answer should be yes, but it is probably extremely hard to prove (and of course impossible to "write down" all the solutions). E.g. for $y=47$ we apparently get the prime number solution $q=\frac{5^{47}-1}{4}$...

Oops, I missed that we only need to find the values of $p$ (instead of finding all solutions).

We define that:$ x=ord_p(a)$ is the smallest positiveinterger satified $a^x\equiv 1\mod p$ It's easy to see that $a^t \equiv 1 \mod p \Leftrightarrow x\mid t$ We have $2^sq=p^y-1=(p-1)(p^{y-1}+...+p+1)$ If y have $\ge 2$ prime divisors difine that is m,n if $r\mid\dfrac{p^m-1}{p-1},r\mid \dfrac{p^n-1}{p-1}$ it's easy to see $m=n=ord_r(p)$ (r is prime number) $\rightarrow wrong$ So y is prime number we also have $\dfrac{p^y-1}{p-1},p-1$ have the same set of prime divisors We have $\gcd(\dfrac{p^y-1}{p-1},p-1)\mid p$ so $p^y-1=p-1$ or $p^y-1=y(p-1)$ $\textbf{Case 1}$$p^y-1=p-1$ it's wrong because $y>1$ $\textbf{Case 2}$$p^y-1=y(p-1)=2^s.q\rightarrow y=2$ or $y=q$ If $y=2\rightarrow y=p+1\rightarrow p=2(unsatisfactory)$ If $y=q\rightarrow p-1=2^s,q\mid p-1\rightarrow y=q=2\rightarrow p^2-1=2^s\rightarrow p=3$

hungt1k31cht wrote: we also have $\dfrac{p^y-1}{p-1},p-1$ have the same set of prime divisors Why?

hungt1k31cht wrote: We define that:$ x=ord_p(a)$ is the smallest positiveinterger satified $a^x\equiv 1\mod p$ It's easy to see that $a^t \equiv 1 \mod p \Leftrightarrow x\mid t$ We have $2^sq=p^y-1=(p-1)(p^{y-1}+...+p+1)$ If y have $\ge 2$ prime divisors difine that is m,n if $r\mid\dfrac{p^m-1}{p-1},r\mid \dfrac{p^n-1}{p-1}$ it's easy to see $m=n=ord_r(p)$ (r is prime number) $\rightarrow wrong$ So y is prime number we also have $\dfrac{p^y-1}{p-1},p-1$ have the same set of odd prime divisors We have $\gcd(\dfrac{p^y-1}{p-1},p-1)\mid y$ so $p^y-1=p-1$ or $p^y-1=y(p-1)$ $\textbf{Case 1}$$p^y-1=p-1$ it's wrong because $y>1$ $\textbf{Case 2}$$p^y-1=y(p-1)=2^s.q\rightarrow y=2$ or $y=q$ If $y=2\rightarrow y=p+1\rightarrow p=2(unsatisfactory)$ If $y=q\rightarrow p-1=2^s,q\mid p-1\rightarrow y=q=2\rightarrow p^2-1=2^s\rightarrow p=3$

Tintarn wrote: hungt1k31cht wrote: we also have $\dfrac{p^y-1}{p-1},p-1$ have the same set of prime divisors Why? sorry that is odd prime divisors

I think this question can be solved using Zsigmondy's theorem, but some anomalies arise.