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

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 $$2$$?

• If $p_1+p_2+p_3 = 2p_1p_2p_3 -17$ subtract $p_3$ from both sides: $p_1 +p_2 = (2p_1p_2-1)p_3-17=(p_1p_2 - 1)p_3 -17+ p_1p_2p_3$. Subtract $p_2$ from both sides: $p_1=(p_1p_2-1)p_3-17+(p_1p_3-1)p_2=(p_1p_2-1)p_3-17+(p_1(p_3-1)-1)p_2 + p_1p_2$. Subtract $p_1$ from both sides: $0=(p_1p_2-1)p_3-17+(p_1(p_3-1)-1)p_2 +(p_2-1)p_1$. But if $p_1,p_2,p_3\ge 3$ then $0=(p_1p_2-1)p_3-17+(p_1(p_3-1)-1)p_2 +(p_2-1)p_1>8*3-17+(3*2-1)*3+2*3=28$. That's not true so you can't have $p_1,p_2,p_3>2$. – fleablood Nov 24 '20 at 0:22

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.

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. $$p_1,p_2,p_3,p_4 = 2,2,2,3$$

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

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

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

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.