Investigating a Prime Pattern

The smallest positive integer $n$ for which the numbers $n^2 + 1$, $n^2 + 3$, $n^2 + 7$, $n^2 + 9$, $n^2 + 13$, and $n^2 + 27$ are consecutive primes is $10$. The sum of all such integers $n$ below one-million is $1242490$.
What is the sum of all such integers $n$ below $150$ million?

This problem is modular arithmetic based and the approach to get to the solution is as follows:

A well-known result about primes is that a prime $p > 3$ gives a remainder of either $1$ or $-1$ when divided by $6$. It means that $p$ can be expressed in one of following two forms:

$p = 6k – 1$ or $p = 6k + 1$, where $k$ is some integer.

Let’s focus first on first two numbers: $n^2 + 1$ and $n^2 + 3$. The difference between them is $2$ which means for these to be consecutive primes, they must be $2$ and $3$. So, either $n^2 + 1 = 2$ and $n^2 + 3 = 3$ or $n^2 + 1 = 3$ and $n^2 + 3 = 2$.

The first case leads to a negative value of $n$ which is not allowed. For the second case, $n = \sqrt{2}$ which is not an integer.

So, $n^2 + 1$ and $n^2 + 3$ cannot be consecutive primes. This also applies to other pairs $(n^2 + 1, n^2 + 3)$, $(n^2 + 7, n^2 + 9)$, and $(n^2 + 13, n^2 + 27)$ for the same reason — their differences are less than $6$ which won’t allow them to be consecutive primes.

Therefore, there are no positive integers $n < 150$ million for which $n^2 + 1$, $n^2 + 3$, $n^2 + 7$, $n^2 + 9$, $n^2 + 13$, and $n^2 + 27$ are consecutive primes. So, the sum of all such integers is $0$.

More Answers: