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

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.

• I only see two pairs that work... $(p,q)=(3,2)$ or $(2,3)$. Do you have other examples? – lulu Jan 12 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 at 17:46
• 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 at 17:54

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.

• 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 at 17:59

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 $$p\equiv 0\mod 3 \implies p = 3$$

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. $$(p,q)=(3,2) , (2,3)$$