Sort of Prime number: ${p(n)=4n^2-4n-1}$

Question

Experiment with quadratic expression $ax^2+bx+c$ motivated by Euler's prime generating polynomial $n^2+n+41$

Given that $\color{red}{p(n)=4n^2-4n-1}$

Conjecture: There is always an integer $n$ value range from $\color{blue}{5k+1}$ to $\color{blue}{5k+5}$, for $k\ge0$, that $p(n)$ will always give at least a prime number.

For examples

$k=0$, $n:1-5$ : [$p(2)=7$]

$k=1$, $n:6-10$ :[$p(7)=167$]

$k=2$, $n:11-15$ :[$p(11)=439$]

$k=100$, $n:501-505$:[$p(502)=1006007$]

So far the conjecture seem correct, but we are trying to find an example where it does not give a prime number but seem not working, hence we naturally ask the question

How does one prove or disprove this conjecture?

Answer 1

The conjecture is false; try $k=7\cdot23\cdot 47\cdot 79=597793$. The $5k+1$ version is the only hard one to test, and turns out to be composite (just 2 prime factors though). The others are multiples of $7,23,47,79$ respectively.

Note: The technique I used was to simplify $$p(5k+2)=4(5k+2)^2-4(5k+2)-1=100k^2+60k+7$$ so if $k$ is a multiple of $7$, so is $p(5k+2)$.

Answer 2

The conjecture is false. Maple gives the smallest counterexample as $k=36$, when the values are $$130319,\ 131767,\ 133223,\ 134687,\ 136159$$ which are all composite.