Do this equation have other solutions other than $P(52)=241$; $3^{5+7+11+13+\cdots+P(n-3)+P(n-2)}=P(n-1) \mod{P(n)}$. Here $P(n)$ is the $n$-th odd prime. I have checked $n$ up to $5500$, but I couldn't find any other solutions
Asked
Active
Viewed 54 times
0
-
possible duplicate of Modularity and prime number sequence – Julián Aguirre Feb 28 '15 at 22:55
-
@JuliánAguirre,no it's different 3,5,7,11 (odd primes only) – Nana J Feb 28 '15 at 22:57
-
You are right. I retracted my vote to close. – Julián Aguirre Feb 28 '15 at 23:01
1 Answers
1
$$ {\huge3}^{\displaystyle\sum_{n=3}^{730690}P(n)}{\Large\equiv P(730691) \mod P(730692)}. $$ $$\begin{align} P(730691)&=11\,067\,689\\ P(730692)&=11\,067\,691 \end{align}$$ There are no more solutions with $n\le10^7$.
Julián Aguirre
- 76,354