For primes, $p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$ I am looking for primes that would satisfy this equation:
$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$
I started by proving that an odd amount of these primes (two or four) cannot be equal to $2$ due to the parity of the left hand side and right hand side.
Then I assumed that exactly three of these primes would be equal to  $2$ and easily found a solution of $(2,2,2,3)$ and its permutations.
My problem now is, I cannot say that there isn't a fitting set of primes, out of which exactly one would be equal to $2$, or for all primes being odd. That leaves me with
$p_1+p_2+p_3=2p_1 p_2 p_3 - 17$
and
$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$
respectively, and I am not sure how to go about those.
Is there a way to solve this assuming one of the primes is equal to $2$ and none of them is equal to  $2$?
 A: Hint: Say $p_4\geq p_3\geq p_2\geq p_1\geq 2$, then we get $$4p_4\geq 8p_4-15$$
A: The left hand side grows linearly in the variables, while the right hand side grows as a degree four polynomial. We can use this to get a bound on the variables.
More precisely:
Without loss of generality assume that $p_1 \le p_2 \le p_3 \le p_4$ (all the other solutions can be obtained by permuting such solutions).
Since $p_4$ is a prime, $p_4 \ge 2$ and hence $8p_4 \ge 16 > 15$.
Now $p_1 p_2 p_3 p_4 = 15 + p_1 + p_2 + p_3 + p_4 < 8p_4 + p_4 + p_4 + p_4 + p_4 = 12 p_4$, and hence $p_1 p_2 p_3 < 12$.
If $p_3 \ge 3$, then we get $12 > p_1 p_2 p_3 \ge 2 \cdot 2 \cdot 3 = 12$, a contradiction. Hence $p_3 \le 2$. But $p_1 \le p_2 \le p_3$, so the only possible solutions is $p_3 = p_2 = p_1 = 2$.
Now the original equation becomes $6 + p_4 = 8p_4 - 15$, which yields $p_4 = 3$.
Clearly $(p_1, p_2, p_3, p_4) = (2, 2, 2, 3)$ is indeed a solution, so the solutions are
$(p_1, p_2, p_3, p_4) \in \{(2, 2, 2, 3), (2,2,3,2), (2,3,2,2), (3,2,2,2)\}$.
A: The question reduces to the following:
Find $p_1,p_2,p_3,p_4$ such that the equation
$$p_1p_2p_3p_4 - (p_1+p_2+p_3+p_4) = 15.$$
We claim however that there aren't too many possibilities to check here. Let us assume WLOG that $p_1 \le p_2 \le p_3 \le p_4$. Then
$$p_1p_2p_3p_4 -(p_1+p_2+p_3+p_4) = p_4(p_1p_2p_3 - \frac{p_1+p_2+p_3}{p_4}-1$$
$$ \ge p_4(8-4-1) = 3p_4.$$
[This because $p_1p_2p_3$ has to be at least 8 because each of $p_1,p_2,p_3$ is at least 2, and as $p_4 \ge p_1,p_2,p_3$, the fraction $\frac{p_1+p_2+p_3}{p_4}$ cannot be more than 3.]
So $3p_4$ cannot be more than 15 which implies that the largest prime $p_4$ cannot be more than 5.
Can you finish from here.
