For primes, $p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

Question

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 $2$?

Answer 1

Hint: Say $p_4\geq p_3\geq p_2\geq p_1\geq 2$, then we get $$4p_4\geq 8p_4-15$$

Answer 2

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

$(p_1, p_2, p_3, p_4) \in \{(2, 2, 2, 3), (2,2,3,2), (2,3,2,2), (3,2,2,2)\}$.

Answer 3

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.

Answer 4

The first thing I think is that $p_1p_2p_3p_4$, which is a multiple of many $p_1$s,$p_2$s, $p_3$s, and $p_4$s, is certainly going to be a lot bigger than $p_1+p_2+p_3+p_4$.

Indeed. $p_1p_2p_3p_4 = (p_1p_2p_3 -1)p_4 + p_4=$

$(p_1p_2p_3-1)(p_4-1) + p_1p_2p_3 -1 + p+4 =$

$(p_1p_2p_3-1)(p_4-1) + (p_1p_2 -1)p_3 -1 + p_3 + p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) +p_1p_2 -2 +p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)p_2 -2 + p_2+p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) + p_1 -3 p_2+p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) -3+ (p_1 + p_2+p_3+p_4)\ge $

$(8-1)(2-1) + (4-1)(2-1)+(2-1)(2-1)-3+ (p_1 + p_2+p_3+p_4)=$

$11+ (p_1 + p_2+p_3+p_4)$.

Now we want $p_1+p_2+p_3+p_4 = p_1p_2p_3p_4 -15 =(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) -3+ (p_1 + p_2+p_3+p_4)-15$

or in other words

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1)=18$.

Note: $p_1p_2p_3-1 \ge 7$ so $p_4-1 \le 2$. And $p_4-1 \ge 1$ so $p_1p_2p_3-1< 17$.

The tells us that $p_4 = 2,3$ and at least two of $p_1,p_2,p_3=2$.

If $p_4 = 3$ the only option is $p_1=p_2=p_3=2$.

If $p_4 =2$ we can have $p_1p_2p_3-1=7, 11$.

We can brute force this out, but as it was arbitrary which terms we factored out first. We can assume without loss of generality that $p_4\ge p_3,p_2,p_1$. So we have either $p_3=3$ and $p_1=p_2=p_3=2$.

Or that $p_3=2$ and all the primes are $2$. Butbut... $2+2+2+2\ne 2*2*2*2-15$ so $(2,2,2,3)$ is the only solution.

==== old answer below ====

If we assume $p_1 \le p_2 \le p_3\le p_4$ then

$p_1+p_2+ p_3 + p_4 = p_1p_2p_3p_4-15$

$15 = p_1p_2p_3p_4 - p_1-p_2-p_3-p_4=$

$p_1(p_2p_3p_4 -1) - p_2 - p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1) + (p_2p_3p_4 -1)- p_2 - p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_3p_4-1)p_2 -1-p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + p_3p_4 - 2 - p_3 - p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_4-1)p_3 -p_4 -2=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1) -3$

So

$12 = (p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1)$

Now $p_2p_3p_4 - 1\ge 7$ so $p_1-1 < 2$ so $p_1-1 =1$ and $p_1 = 2$.

And $p_2p_3p_4-1 \ge 7$ so

$5 \ge (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1)$ and $p_3p_4-1 \ge 3$ so $p_2-1 < 2$ and $p_2 = 2$.

And as $p_3p_4-1 \ge 3$ we have

$2\ge (p_3-1)(p_4-1)$.

If $p_3-1\ge 2$ then $p_4-1 \ge 2$ and that'd imply $2\ge 4$ which is a contradiction, so $p_3-1=1$ and $p_3 = 2$

And with that we have

$2+2+2+p_4 = 8p_4 -15$

$21 = 7p_4$

$p_4 =3$.

So that's the only (up to order) solution. $p_1,p_2,p_3,p_4 = 2,2,2,3$