1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I only see two pairs that work... $(p,q)=(3,2)$ or $(2,3)$. Do you have other examples? $\endgroup$ – lulu Jan 12 at 17:41
  • $\begingroup$ 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. $\endgroup$ – Silverleaf1 Jan 12 at 17:46
  • 1
    $\begingroup$ 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. $\endgroup$ – fleablood Jan 12 at 17:54
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $2(p+1)\equiv -p-1$. $\endgroup$ – fleablood Jan 12 at 17:59
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.