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

Mathematics Asked by vlnkaowo on December 12, 2020

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 $2$?

4 Answers

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 $p_4ge p_3,p_2,p_1$. 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... $2+2+2+2ne 2*2*2*2-15$ so $(2,2,2,3)$ is the only solution.

==== old answer below ====

If we assume $p_1 le p_2 le p_3le p_4$ 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 $p_2p_3p_4 - 1ge 7$ 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

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

If $p_3-1ge 2$ then $p_4-1 ge 2$ and that'd imply $2ge 4$ 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. $p_1,p_2,p_3,p_4 = 2,2,2,3$

Correct answer by fleablood on December 12, 2020

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.

Answered by Mike on December 12, 2020

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

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

Answered by Chaitanya Tappu on December 12, 2020

Hint: Say $p_4geq p_3geq p_2geq p_1geq 2$, then we get $$4p_4geq 8p_4-15$$

Answered by Aqua on December 12, 2020

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