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Find all prime numbers $p$ and $q$, such that $7p+q$ and $pq+11$ are also prime numbers.

Based on the fact that all primes, besides 2, are odd, I found that either $p$ or $q$ must be $2$ in order for $pq+11$ to be a prime number. From here, I found several pairs of $p$ and $q$ that work, but I don't know how to find all $p$ and $q$. I tried letting

$7p+q=r$

$pq+11=s$

and then adding the equations and using SFFT to get:

$(p+1)(q+7)=r+s-4$

but it doesn't really help.

Silverleaf1
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  • I only see two pairs that work... $(p,q)=(3,2)$ or $(2,3)$. Do you have other examples? – lulu Jan 12 '20 at 17:41
  • i thought i found another pair of p and q that work but turns out it doesn't. I think 2 and 3 are the only two pairs. – Silverleaf1 Jan 12 '20 at 17:46
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    I think lulu's shown those are the only pairs. Either $p=2$ and $14+q\equiv q-1\pmod 3$ and $2q +11\equiv -q-1\pmod 3$. But unless $q=3$ or $3|q$ then one one of those equations is equiv $0\pmod 3$. Similarly if $q=2$ then $7p +2\equiv p-1\pmod 3$ and $2p+11\equiv -p-1\pmod 3$ gives the same conclusion. – fleablood Jan 12 '20 at 17:54

3 Answers3

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So, say $p=2$. Then your expressions are $14+q$ and $2q+11$. Working $\pmod 3$ we see that these are $q-1$ and $2-q$ Easy to see that one of these is divisible by $3$ unless $q=3$ which is a valid example.

Now say $q=2$, Then your expressions are $7p+2$ and $2p+11$. Working $\pmod 3$ we see that these are $p+2$ and $2(p+1)$ and again one of these terms must be divisible by $3$ unless $p=3$, which is again a valid example.

lulu
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  • It might be illuminating to the OP (or not) that as $2\equiv -1\pmod 3$ that the equivalences generated by $p=2$, that is $q-1$ and $2-q\equiv -q-1$, turn out to be the exact same as the equivalences generated by $q=2$, that is $p+2\equiv p-1$ and $2(p+1)\equiv -p-1$. – fleablood Jan 12 '20 at 17:59
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We must have either $p=2$ or $q=2$ , Otherwise $7p+q$ is even .

If $p=2$ , then let $14+q=x$ and $2q+11 = y$. Adding these equation and taking $\text{modulo } 3$ , we get :

$$x+y \equiv 1\mod 3 \implies x,y \equiv 2\mod 3$$

Substituting this back into the original equation gives $$q\equiv 0\mod 3 \implies q = 3$$

If $q = 2$ , then let $7p + 2 = x$ and $2p + 11 = y$ . Again adding and taking $\text{ modulo } 3$ , we get :

$$x+y\equiv 1\mod 3\implies x,y \equiv2 \mod 3$$

Substituting this back into the original equation gives $$p\equiv 0\mod 3 \implies p = 3$$

The Demonix _ Hermit
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As you have already pointed out that $p=2$ or $q=2$ Take cases , $Case$ $1$. Let, $p=2$ , Since we know every prime greater than $3$ is of the form $6k+1$ or $6k-1$. Taking $q=6k+1$ implies $7p+q$ divisible by $3$ and taking $q=6k-1$ implies $pq+11$ is divisible by $3$. Thus only possible $q=3$. In that case both expressions are equal to $17$.

$Case$ $2$. Let, $q=2$ if $p=6k+1$ then $7p+q$ is divisible by $3$ and for $p=6k-1$ then $pq+11$ is divisible by $3$. Thus only possible $p=3$. Then clearly first expression is $23$ and second expression is $17$ and both of these are primes. So only solution. $(p,q)=(3,2) , (2,3)$