Find all primes $(p,q)$ such that $p|q+6$ and $q|p+7$

Question

Find all primes $(p,q)$ such that $p|q+6$ and $q|p+7$

I haven't found any. I initially started from $p,q\gt 3$ since, by a simple substitution you get, if of $p=2$ then $q|8$ and $q=2$, but then $2|9$ and it's a contradiction. Similarly happens with 3. Then, $p,q$ are odd and greater than 3. Also, I tried using $p,q\equiv \pm1\pmod 6$ and linear combinations, but I haven't gotten anything and I don't know how to proceed.

I would prefer a suggestion rather than an answer, if possible without congruences, thanks beforehand.

Answer 1

We have \begin{align} q\mid p+7& \implies 2\nmid q, 2\nmid p \implies 2\mid |p-q|\\ q-7\leq p \leq q+6 &\implies -7\leq p-q \leq 6 \implies 0\leq q+6-p\leq 13. \end{align} Since $2\mid q+6-p$ and $p\mid q+6-p$, we have the following cases: $$\begin{cases} q+6-p = 0& \implies p = q+6 \implies q\mid q+13 \implies q\mid 13 \implies q =13\implies p=19 \\ p = 3, q+6-p = 6 &\implies q= 3 \implies q\nmid p+7\text{ i.e, no solution} \\ p = 3, q+6-p = 12&\implies q = 9 \text{ i.e, no solution} \\ p = 5, q+6-p = 10&\implies q=9 \text{ i.e, no solution.} \end{cases} $$ Therefore, the only solution is $(19,13)$.

Answer 2

Hint and Partial answer.- You need $$\begin{cases}q+6=px\\p+7=qy\end{cases}\Rightarrow\begin{cases}p=\frac{7+6y}{xy-1}\\q=\frac{6+7x}{xy-1}\end{cases}$$ You can take successively $xy=2,3,4,\cdots,n$ but since the first numerator is always odd you can reduce to the values $xy=2,4,6,\cdots, 2n$.

For the first $xy=2$ you have the only possibilities $(x,y)=(1,2).(2,1)$ from which only $(x,y)=(1,2)$ works giving the solution $(p,q)=(19,13)$.

it is clear that the possibilities for to have numerators greater than the denominators are restricted so we have to look at some few values. I feel that $(19,13)$ could be the only solution or if there is another then there are only few of them.

Answer 3

Since $p\,|\,q+6$ we have $q+6≥p$.

Since $q\,|\,p+7$ we have $p+7≥q$ or, equivalently $p≥q-7$

We combine these to get $$q+6≥p≥q-7$$

We now distinguish two cases:

Case I: $q+6>p$

Then, since $p$ is a divisor of $q+6$ we must have $\frac {q+6}2≥p$. Thus $$\frac {q+6}2≥q-7\implies q+6≥2q-14\implies 20≥q$$

Thus we only have to check $q\in \{2,3,5,7,11,13,17,19\}$ with $p≤\frac {q+6}2$. By trial and error, there are no solutions of this type.

Case II: $q+6=p$.

Then we must have $p+7>q$ whence $q≤\frac {p+7}2$ or $2q-7≤p=q+6$ so $q≤13$. We check the various cases and see that $(p,q)=(19,13)$ is the only solution of this type.

Answer 4

From $q\mid p+7$ we have $p+7=qk$ for some positive integer $k$.

• If $k=1$ then $p+7=q$ so $p,q$ one is even, so $p=2$ and $q=9$. Not good.
• If $k=2$ then $p+7=2q$ and since $p\mid 2q+12$, then $p\mid 19$, so $p=19$ and $q=13$.
• If $k=3$ then $p+7=3q$ so $p,q$ one is even, so $p=2$ and $q=3$ which doesn't work in second relation.
• If $k\geq 4$ then $p+7\geq 4q$. But $p\leq q+6$ so $3q\leq 13\implies q\leq 3$. If $q=2$ then $p=2$ which doesn't work and if $q=3$ then $p= 3$ which also doesn't work.

Conclusion: $p= 19$ and $q=13$.

Answer 5

Although late, here is another proof.

First of all, let $q+6 = pk_1,\ p+7=qk_2$, for some $k_1,k_2\ge 1$. Note that $p\ne q$, because otherwise, $p(k_2-k_1)=1\implies p=1$, which is not a prime number. Now as $p|q+6$, clearly, $p\ne 2,3$ as $p\ne q$. Now, let $q>p$. Then we have $p+7>k_2p\implies (k_2-1)p<7$. Since $p\ge 5$,this implies $k_2=1$. However, $p+7=k_2q=q\implies$ $q$ is even (and $\ge 2$), which is not possible. Therefore, $k_2\ge2$, and we must have $p>q$, i.e., $q+6>k_1q\implies (k_1-1)q<6$, which implies that either $k_1=1$, or $q=2$ or $3$. However, as $p\ge 5$, and $p|q+6$, we cannot have $q=2$ or $3$. Therefore, $k_1=1$.

Thus, we have, $q+6=p,\ p+7=2q\implies, p=19,\ q=13$.