Prove $(c^2-1)/4 \pm 1$ is not divisible by $2, 3, $ or $5$ if $c>3$ is an odd prime

Question

In this question asked the other day, RTn conjectured that, if $c>3$ is an odd prime,

then $k=\dfrac{c^2\pm4-1}4=\dfrac{c^2-1}4\pm1$ is prime. As the answers and comments there show,

this conjecture is not true; some numbers of that form are multiples of $7$, $11$, $13$, etc.

I suspect, however, that RTn was misled because $k$ cannot be divisible by $2$, $3$, or $5$.

This question is how to prove this.

I'm taking the option of answering my own question, but I'd appreciate feedback or other answers.

Answer 1

If $c>3$ is an odd prime, then $c$ is not divisible by $2, 3, $ or $5,$

unless $c=5, $ in which case the claim can be easily verified.

If $c$ is not divisible by $2,$ then $c^2-1=(c+1)(c-1)$ is the product of two consecutive

even numbers so divisible by $8,$ so $\dfrac {c^2-1}4$ is even, so $\dfrac{c^2-1}4\pm1$ is not divisible by $2$.
If $c$ is not divisible by $3$, then $c+1$ or $c-1$ is, so $c^2-1$ is, so $\dfrac{c^2-1}4$ is (since $4\equiv1\mod3),$

so $\dfrac{c^2-1}4\pm1$ is not.

Finally, if $c$ is not divisible by $5,$ then $c^2\equiv 1$ or $4\mod5$, so $c^2-1\equiv 0$ or $3\mod 5$;

since $4\equiv-1\mod 5,$ that means $\dfrac{c^2-1 }4\equiv0 $ or $2\mod 5$, so $\dfrac{c^2-1}4\pm1\not\equiv0\mod5.$

Answer 2

It is easy to characterize all prime divisors $p > 5$.

$\!\!\bmod p\!:\,\ 0\equiv (x^2-1)/4\pm 1\iff x^2\equiv \mp4+1 \equiv \color{#c00}{-3},\color{#0a0}5$

By quadratic reciprocity $\,(\color{#c00}{-3}|p) = 1\!\iff\! p\equiv1\pmod{\!3};\,$ $\,(\color{#0a0}5|p)=1\!\iff\! p\equiv\pm1 \pmod{\!5}$

It follows by CRT that the prime divisors $p>5$ are exactly those primes $\not\equiv 2,8\pmod{\!15},\,$ i.e. those primes $\equiv 1,4,7,11,13,14\pmod{\!15}$. Hence - further - it is never divisible by these primes $\, 17,23,47,53,83,107,113,137,167,173,197,\ldots$