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$p, q, r$ are prime numbers such that $$pqr = 19 (p + q + r)$$

We have to find the value $\sqrt{p} + \sqrt{q} + \sqrt{r}$.

I could not solve the problem and sought a method of my own. It seems that by putting $p = 19$ and evaluating further, a solution may arise.

I need a general solution without putting any specific value for given variables. Since I just started studying number theory, give me sufficient hint.

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  • $\begingroup$ Shoudn't it be $p+q+r$ instead of $p+q++$? $\endgroup$ – José Carlos Santos Jun 15 '17 at 8:16
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The point is that since $19$ is a factor of $19(p+q+r)$, it must also be a factor of $pqr$. So you know that either $p=19$, $q=19$ or $r=19$. It doesn't actually make a difference which one (since you can reorder $p,q,r$ and you don't change the answer), so it may as well be $p$.

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There is no problem with taking $p=19$. Since $pqr$ is a multiple of $19$ and, since $p$, $q$, and $r$ are primes, then one of them must be $19$. There is no loss in generality assuming that $p=19$.

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The problem is symmetric in $p,q,r$ hence you can set wlog $p=19$. At this point you have $19qr=19(19+q+r)$, i.e., $$ q=\frac{r+19}{r-1}=1+\frac{20}{r-1}. $$ Hence $r=3$ and $q=11$ (or viceversa), and the solution is $$ \sqrt{3}+\sqrt{11}+\sqrt{19}. $$

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