A Prime Quadratic

Relevant wiki: Quadratic Diophantine Equations - Problem Solving

Only the primes $2$ and $5$ appear seven or more times.

$(2,5,2),(2,5,3),(2,7,5),(2,11,5)$ and their rearrangements show the fact that $2$ and $5$ appear more than seven times.

We will show that if either $l=3$ or $l$ is a prime greater than $5$ then $l$ occurs at most six times as an element of a triple in $S$ . ) Note that $(p,q,r) \in S$ if and only if $q^2-4pr=a^2$ for some integer $a$ . In particular, $4pr \geq 16$ this forces $q \geq 5$

• If $q$ is odd then $a$ is odd and so $q^2 \equiv a^2 \equiv 1 \pmod{8}$

Consequently, one of $p,r$ must equal $2$ .

If $r=2$ then $8p=q^2-a^2=(q+a)(q-a)$ .Since both factors are of same sign and their sum is $q>0$ ,both factors are positive. Since they are also both even we have $q+a \in \{2,4,2p,4p\}$ and so $q \in {2p+1,p+2}$ . Similarly if $p=2$ then $q \in {2r+1,r+2}$ . Consequently, $l$ occurs at most twice as there are prime numbers in the list $\{2l+1, l+2,\frac{l-1}{2},l-2\}$ For $l=3$ , $2l-1=1$ is not a prime. For $l \geq 7$ the numbers $l-2,l+2$ cannot all be prime, since one of them is always a multiple of $3$ .

Remark The above argument shows that the cases listed for the $5$ are the only cases that can occur. By the contrast there are infinitely many cases where $2$ occurs if either the twin prime conjecture holds or there are infinitely many Sophie Germain primes. .

We have quadratic equation $px^2 + qx + r = 0$ . Using the rational root theorem we know that only $\pm 1, \ \pm r, \ \pm \dfrac1p, \ \pm \dfrac rp$ can be rational solutions to the equation.

Now we get $8$ equations by substituting the possible roots to the equation.

$p - q + r = 0 \\ p + q + r = 0 \\ pr^2 - qr + r = 0 \\ pr^2 + qr + r = 0 \\ \dfrac{1}{p} - \dfrac{q}{p} + r = 0 \\ \dfrac{1}{p} + \dfrac{q}{p} + r = 0 \\ \dfrac{r^2}{p} - \dfrac{qr}{p} + r = 0 \\ \dfrac{r^2}{p} + \dfrac{qr}{p} + r = 0$

Since the primes are positive, after some simplifications we are left with $2$ equations.

$p + r = q \\ pr + 1 = q$

If $q = 2$ the equations have no solutions, hence $q$ is odd. This means that either $p = 2$ or $r = 2$ .

$2 + r = q \\ 2r + 1 = q \\ 2 + p = q \\ 2p + 1 = q$

Since for each odd prime $n$ there can be only $1$ corresponding twin prime if $n \neq 5$ (because $5$ is the middle number in the only prime triple) and only $1$ corresponding Sophie Germain Prime and for each Sophie Germain Prime there can be only $1$ corresponding prime, $n = 5$ because otherwise $n$ can only appear in $6$ elements of $S$ .

We can easily find that there are indeed $7$ elements containing $5$ in $S$ and since all of those elements also contain $2$ , the only primes which appear enough times are $2$ and $5$ and hence the answer is $\boxed{7}$ .