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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?

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    $\begingroup$ Personally I think this is a perfectly adequate solution. $\endgroup$
    – Luke Collins
    Commented Nov 23, 2020 at 21:03
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    $\begingroup$ Why can you conclude that both of $p_4,p_5$ are $\ge 5$? $p_4=2$, $p_5=17$ seems to work fine. $\endgroup$
    – Hagen von Eitzen
    Commented Nov 23, 2020 at 21:07
  • $\begingroup$ @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? $\endgroup$
    – user635053
    Commented Nov 23, 2020 at 21:12
  • $\begingroup$ reminds me of this question $\endgroup$
    – J. W. Tanner
    Commented Nov 23, 2020 at 21:29

2 Answers 2

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

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  • $\begingroup$ If the condition distinct primes included in the problem then the solution is unique. $\endgroup$
    – Lawrence Mano
    Commented Nov 23, 2020 at 22:19
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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.

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