All primes $p, q$ that satisfy $p \cdot q=3p +7q$

Question

Suppose I would want to find all pars of primes $p,q$ of the form $p \cdot q=3p +7q$, observe that $$p \cdot q-7q=3p.$$ $$q(p-7)=3p.$$ Now I am kind of stuck, should I do a case distinction on $p$ and $q$... or...

Answer 1

The RHS of $q(p-7)=3p$ is a product of two primes. Therefore the LHS must also be. Then we have the alternatives $q=3$ or $q=p$. The case $q=3$ leads to $p-7=p$ which is impossible. The case $q=p$ leads to $p-7=3$ i.e. $p=10$, which is not a prime. Thus there are no prime solutions.

Answer 2

Hint: Write $$p=7+\frac{21}{q-3}$$

Answer 3

Consider the parity of $p$ and $q$. Suppose both $p$ and $q$ are odd. Then, we have that $pq$ is odd. However, we also then have that $3p\equiv 1\pmod 2$ and $7q\equiv 1 \pmod 2$, which gives $3p+7q\equiv 0\pmod 2$, which is even. So, $p$ and $q$ cannot both be odd, at least one must be even.

Since $p$ and $q$ are primes, and there is exactly one even prime, at least one out of $p$ or $q$ must be $2$. You can now substitute each case to find the solutions, if they exist.

Subsituting $p=2$ gives $5q=-6$, and substituting $q=2$ gives $p=-14$. So, there are no solutions.

Answer 4

If $p \neq 7$ and $p \neq q$ then $p$ would divide the LHS but not the RHS leading to a contradiction . Hence, $p=7$ or $p=q$. If $p=q$ the equation becomes $p=10$, a contradiction.$p=7$ also leads to a contradiction, so there are no primes satisfying this equation.

Answer 5

No (positive or negative) primes $p$ and $q$ work. Write $$(p-1)(q-1)=2p+6q+1\,.$$ Thus, $(p-1)(q-1)$ is odd, so $p$ and $q$ are even. That is, $$p,q\in\{-2,+2\}\,.$$ Hence, $$|pq|=4\text{ but }|3p+7q|\geq 3(-2)+7(2)=8>4\,,$$ so the equality $pq=3p+7q$ does not hold.

However, there are integer solutions. Write $$(p-7)(q-3)=21\,.$$ Then, there are $8$ solutions $(p,q)\in\mathbb{Z}\times\mathbb{Z}$: $$(8,24)\,,\,\,(10,10)\,,\,\,(14,6)\,,\,\,(28,4)\,,$$ $$(6,-18)\,,\,\,(4,-4)\,,\,\,(0,0)\,,\text{ and }(-14,2)\,.$$

Answer 6

You can rearrange the equation as $(p-7)(q-3)=21$. The factors must both be odd and this only happens if $p$ and $q$ are both even. And $p=\pm 2$ or $q=\pm 2$ don't work.

Note that this kind of thing is often useful where you have an expression in $pq$ equal to a linear combination of $p$ and $q$

Answer 7

$p$ and $q$ can't be both odd, since then left side is od and right is even.

Case 1. $p=2$ so $q=-6/5 $, impossible.

Case 2. $q=2$ so $p= -14$, impossible.

So no solution.