Three Primes

I will share a slightly different solution than @Mursalin Habib which may be easier than the one he posted.

---

First, just like Mursalin, we subtract $2pqr$ from both sides to get $p^2q^2-2pqr+r^2=pq+r$ This can be factorized as $(pq-r)^2=pq+r$

Now we subtract $pq-r$ from both sides to get $(pq-r)(pq-r-1)=2r$ Since $pq-r$ and $pq-r-1$ are consecutive integers, then either $pq-r=2$ or $pq-r-1=2$ .

Seeing that $pq-r=2$ forces $pq-r-1=1$ and $r=1$ , and since $r$ is prime, this case cannot be the right one. Thus, $pq-r-1=2$ , $pq-r=3$ , and $r=3$ .

This means $pq=6$ , which forces $(p,q)=(2,3)\text{ or }(3,2)$ .

Thus, $p+q+r=2+3+3=\boxed{8}$ .

Let's take a look at what we have here:

$p^2q^2+r^2=2pqr+pq+r \cdots (1)$

And we have to find out $p+q+r$ .

Equation $(1)$ doesn't give us much information about $p$ , $q$ and $r$ in the face of it. But if you move around the terms a little, you get something like this:

$(pq-r)^2=pq+r$ .

This equation certainly looks better than $(1)$ . So that's a start.

After that I rewrote it as:

$\dfrac{pq+r}{pq-r}=pq-r$ .

Then I subtracted $1$ from both sides to get the following equation:

$\dfrac{2r}{pq-r}=pq-r-1 \cdots (3)$ .

Now take a look at the fraction on the left hand side of the equation. This fraction has to be an integer. So, $r$ must divide $pq$ . Since $p$ , $q$ and $r$ are all primes, it means either $p$ or $q$ must be equal to $r$ .

Without loss of generality, let $p =r$ . Then $(3)$ becomes:

$\dfrac{2}{q-1}=r(q-1)-1\cdots (4)$ .

If you again take a look at the fraction on the left hand side, it becomes quite clear that either $q=2$ or $q=3$ . Time for case-checking.

Case 1 : [ $q=2$ ]

Plug $q=2$ into equation $(4)$ . That gives us $r=p=3$ . And $p+q+r=3+2+3=\boxed{8}$ .

Case 2 : [ $q=3$ ]

Plug that in $(4)$ and you're going to get $r=1$ and $1$ is not prime number. So, we disregard this case.

And our final answer is $\boxed{8}$ .

We can rewrite the equation as $(pq)^2 - pq(2r+1) + (r^2-r) = 0$ . Now the Quadratic Formula tells us:

$pq = \frac{2r+1 \pm \sqrt{8r+1}}{2}$

Therefore $8r+1$ is a perfect square. Say it is $k^2$ for some integer $k$ . Then $8r = k^2 - 1 = (k+1)(k-1)$ .

$Lemma:$ I say $r \le 10$ .

Suppose, to the contrary, that $r \ge 11$ . Since $r$ is prime and it divides $(k+1)(k-1)$ , then it divides $k+1$ or $k-1$ (or both). In either case, it must be that $k+1 \ge r$ . Hence $k-1 \ge r-2$ . Then $(k+1)(k-1) \ge r(r-2) \ge 9r$ . This is a contradiction since $(k+1)(k-1) = 8r$ .

Thus $r \le 10$ , and we need $r$ to be prime such that $8r+1$ is a perfect square. The only value which satisfies these conditions is $r=3$ . By the Quadratic Formula, this means $pq = 1$ or $pq = 6$ . Clearly it must be the latter, so that $p,q$ are $2$ and $3$ in some order.

Thus $p+q+r=8$

$p^2q^2+r^2=2pqr+pq+r~\Longrightarrow (pq-r)^2=pq+r~\Longrightarrow \dfrac{2r}{pq-r}=pq-r-1$ Now $pq-r\mid 2r$ so $pq-r$ can take only four values, namely $1,2,r,2r$ . Substituting all one by one and then solving for $r$ shows that $pq-r=r$ because all other values yield non-prime values of $r$ . Using $pq-r=r$ and solving for $r$ gives $r=3$ . Note that $pq-r=r$ implies $pq=2r=6=2\times 3$ . So $p=2, q=3$ WLOG. And hence, $p+q+r=\fbox{8}$ .

p^2-q^2-pq(2r+1)+r(r-1)=0

so we get pq=((2r+1)+sqrt(8r+1))/2; since pq is real, 8r+1 is a perfect square; give trial and error: 8r+1=1 ;r=0 not a prime

8r+1=4;r=3/8 not a prime

8r+1=9;r=1 not a prime

8r+1=16;r=15/8 not a prime

8r+1=25;r=3 gives pq=6; so again by trial method we get p=2 q=3;

so p+q+r=2+3+3=8