11
$\begingroup$

I noticed something as I was playing around with prime numbers. By denoting $p_i$ the $i^{\text{th}}$ prime number, I discovered the following:

$$ \begin{align}\prod_{i=1}^2\left(p_i^{ \ 2}+i\right)&= 28^2-27^2 \\ \prod_{i=1}^5\left(p_i^{ \ 2}+i\right)&=3207^2-27^2.\end{align} $$

Given the general equation

$$\prod_{i=1}^m\left(p_i^{ \ 2}+i\right)=n^2-27^2\tag*{$\big(\left(m,n\right)\in\mathbb{N}^2\big)$}$$

I have tested for $5< m\leqslant 32$ and it appears that for $m > 5$, there does not exist such $n$. I know that $n^2-27^2=(n+27)(n-27)$, but this does not help me make much progress.

Is this true? Can it be verified? Thus far, it remains a conjecture that the only solutions $(m,n)$ are $(2,28)$ and $(5,3207)$.


Source of Testing:

$$\text{Alpertron $-$ Integer Factorization Calculator.}$$


Thank you in advance.

$\endgroup$
  • 3
    $\begingroup$ Verified for $m \leq 50000$.Though I am not sure if there is something special about this, since the squares are simply more and more rare when we go to bigger numbers, so it is just expected, kind of... $\endgroup$ – Sil May 20 '18 at 8:02
  • 1
    $\begingroup$ So, is it $(p_i^2+1)$ or $(p_i^2+i)$ in the general equation? $\endgroup$ – rtybase May 20 '18 at 12:18
  • 2
    $\begingroup$ I can approve that upto $m=10^4$, no further solution exists. I am currently running the verification upto $m=10^5$ $\endgroup$ – Peter May 20 '18 at 16:48
  • 1
    $\begingroup$ @Sil An even "easier" question : Are there infinite many primes of the form $n^2+1$, where $n$ is an integer ? This is an open problem! $\endgroup$ – Peter May 20 '18 at 16:50
  • 2
    $\begingroup$ Don't worry, it was with $i$ :) @Peter Yea, I am afraid number theory is full of those :S $\endgroup$ – Sil May 20 '18 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.