Sum of $5$ primes is $105$ times less than their product I am trying to find a "legitimate" solution to this problem. For primes $p_n$:
$105(p_1+p_2+p_3+p_4+p_5) = p_1 p_2 p_3 p_4 p_5$
I've figured out that it's the best to start with breaking the $105$ down to prime factors: $105 = 3 \times 5 \times 7$, by which I know that $p_1 = 3, p_2 = 5, p_3 = 7$.
With that out of the way, I have
$105(15+p_4+p_5) = 105p_4 p_5$, and therefore $15 + p_4 +p_5=p_4 p_5$.
It's obvious that $p_4 p_5 > 15$, therefore $p_4, p_5 ≥ 5$.
And this where I am stuck. My only idea was to consider that the right hand side is increasing faster than the left hand side, and therefore $p_4,p_5$ cannot be big numbers, and start testing numbers starting from $p_4 = 5, p_5 = 5$, which coincidentally happened to be the solution.
Is there a more legitimate way to solve this problem, different from what I did here?
 A: Starting from
$$ 15+p_4+p_5=p_4p_5,$$
we transform to
$$ (p_4-1)(p_5-1)=p_4p_5-p_4-p_5+1=16.$$
From the known factorizations of $16$, we find that $(p_4,p_5)$ is one of $(2,17)$, $(3,9)$, $(5,5)$, $(9,3)$, $(17,2)$. After removing composites and up to symmetry, we are left with $(2,17)$ and $(5,5)$. So in total, we have (up to permutation) two solutions
$$(2,3,5,7,17)\qquad\text{and}\qquad  (3,5,5,5,7). $$
A: To formalize your idea that $p_4$ and $p_5$ cannot be big numbers, one way to proceed would be as follows.
Without loss of generality, assume that $p_4 \le p_5$. Then, from
$$p_4p_5 = 15 + p_4 + p_5,$$
divide both sides by $p_5$ to get
$$
\begin{align}
p_4 &= \frac{15}{p_5} + \frac{p_4}{p_5} + 1\\
&\le \frac{15}2 + 1 + 1,\\
\end{align}$$
which implies that $p_4$ can be no larger than $7$. You can then see what happens when you set $p_4 = 2, 3, 5, 7$.
A: Your equation can be rewritten
$$p_4 p_5 - p_4 - p_5 - 15 = 0$$
or
$$p_4 p_5 - p_4 - p_5 +1 = 16$$
or
$$(p_4-1)(p_5-1) = 16.$$
Assume $p_4 \leq p_5$, so $p_4 = 1, 2,$ or $4$.  This gives you all the solutions.
