# 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?

• Personally I think this is a perfectly adequate solution. Nov 23, 2020 at 21:03
• Why can you conclude that both of $p_4,p_5$ are $\ge 5$? $p_4=2$, $p_5=17$ seems to work fine. Nov 23, 2020 at 21:07
• @HagenvonEitzen I see, not considering 2 was definitely a mistake here. That still doesn't tell me much about how to go about exercise. Should I then just prove that $p_4, p_5$ can't be both 2, then suppose one of the primes is 2 and solve, and then suppose the primes are odd and solve?
– user635053
Nov 23, 2020 at 21:12
• reminds me of this question Nov 23, 2020 at 21:29

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 $$(2,3,5,7,17)\qquad\text{and}\qquad (3,5,5,5,7).$$

• If the condition distinct primes included in the problem then the solution is unique. Nov 23, 2020 at 22:19

$$p_4 p_5 - p_4 - p_5 - 15 = 0$$
$$p_4 p_5 - p_4 - p_5 +1 = 16$$
$$(p_4-1)(p_5-1) = 16.$$
Assume $$p_4 \leq p_5$$, so $$p_4 = 1, 2,$$ or $$4$$. This gives you all the solutions.