All primes $p, q$ that satisfy $p \cdot q=3p +7q$ 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... 
 A: Hint: Write
$$p=7+\frac{21}{q-3}$$
A: 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.
A: $p$ 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.
A: 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 $(p,q)\in\mathbb{Z}\times\mathbb{Z}$:
$$(8,24)\,,\,\,(10,10)\,,\,\,(14,6)\,,\,\,(28,4)\,,$$
$$(6,-18)\,,\,\,(4,-4)\,,\,\,(0,0)\,,\text{ and }(-14,2)\,.$$
A: 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 $q$
A: 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.
A: 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. 
