SMO Primes!

without loss of generality $p=19$ becasuse $19|pqr$ also can be $r$ o $q$ .

$qr=19+q+r$

$(q-1)(r-1)=20$

$q-1=1,2,4,5,10,20) \Rightarrow q$ can be $2,3,5,11$

If $q=2 \Rightarrow r=21$ so $r$ is not a prime.

If $q=3 \Rightarrow r=11$ so $r$ is a prime. $\star$

If $q=5 \Rightarrow r=6$ so $r$ is not a prime.

If $q=11 \Rightarrow r=3$ so $r$ is a prime. $\star$

So $p=19, q=3, r=11 \therefore \boxed{p^2+q^2+r^2=491}$

First note that the expression is completely symmetric. Since $(p,q,r)\in\mathbb P^3$ , we let $p=19$ . Then we get $qr=19+q+r$ . Rearranging and factorizing gives us $(q-1)(r-1)=20=2\times 10$ . So we have a pair $(q,r)=(3,11)$ of primes. Thus the answer is $3^2+11^2+19^2=\fbox{491}$ .

$pqr = 19(p + q + r)$

Either one of $p$ , $q$ or $r$ is equal to $19$ , given all of them are prime numbers.

WLOG, let's assume $p = 19$ . Then:

$qr = 19 + q + r$

$qr - q = 19 + r$

$q = \frac {19 + r}{r - 1} = 1 + \frac {20}{r - 1}$

Let's test a few values of $r$ to determine $q$ .

$r = 2 \rightarrow q = 21$ ; untrue because $q$ is not prime.

$r = 3 \rightarrow q = 11$ ; valid solution; both $r$ and $q$ are prime.

$r = 5 \rightarrow q = 6$ ; untrue because $q$ is not prime.

$r = 7 \rightarrow q = \frac{13}{3}$ ; untrue because $q$ is not prime, let alone an integer.

$r = 11 \rightarrow q = 3$ ; valid solution, symmetric to the previous one we found with $r = 3$ .

$r = 13 \rightarrow q = \frac{8}{3}$ ; untrue because $q$ is not prime, let alone an integer.

$r = 17 \rightarrow q = \frac{8}{3}$ ; untrue because $q$ is not prime, let alone an integer.

$r = 19 \rightarrow q = \frac{8}{3}$ ; untrue because $q$ is not prime, let alone an integer.

For any $r$ greater than $21$ we don't even bother looking for a solution because then $q$ would not be an integer.

Thus, any solution of the equation given the conditions is a permutation of $(3, 11, 19)$ . Thus, the sum asked equals: $3^{2} + 11^{2} + 19^{2} = 9 + 121 + 361 = 491$ .

Simple sol:

$pqr=19(p+q+r)$

p=19

$qr=19 + q + r=$

$(q-1)(r-1)=20$

q,r are $\le$ 19

both can only be:

{3,5,7,11,13,17,19}

so

$(q-1)$ would be only {2,4,6,10,12,16,18}

if q=3 then r=11 both are primes doing this with the other possible results we end up with only {19,11,3}

so result would be $19^{2}+11^{2}+3^{2}=491$

p should be 19, without loss of generality. Let q=2 m+1 and r=2 n+1 (Neither q nor r can be equal to 2 - the only even prime number, that can be easily checked). Putting the above variables into the given equation results in m*n=5. Considering m and n are positive whole numbers, without the loss of generality we end up with m=1 and n=5, which gives us q=3 and r=11.

since RHS is amultiple of 19 so lhs should also contain 19 and p,q,r are primes then one of p,q or r should be 19 as 19 is also a prime. Let r be 19 so we are left with (q,p) which can be either (11,3) or (3,11) so p=3 or 11,q=11 or 3 and r=19

So sum of squares =491

Used Octave (MatLab) to brute force my way into an answer:

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a = primes(500);

for i = 1:75

    for j = 1:75

        for k = 1:75

            p = a(i);
            q = a(j);
            r = a(k);

            t1 = p*q*r;
            t2 = 19*(p+q+r);

            if t1==t2
                p
                q
                r

                flag=1;

                break
            end

            if flag==1
                break
            end
        end

        if flag==1
            break
        end
    end

    if flag==1
        break
    end
end

(p^2)+(q^2)+(r^2)

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OUTPUT:

p =  3

q =  11

r =  19

ans =  491

Would be rather simple to make on TI-basic, too. In this case, i, j, and k can be extended to 95 (not just the 75 they are at now) since there are 95 primes from 1 to 500. I figured that it was already overkill, but with a little patience this could have been used to find much higher primes.

Just wanted to mention that since there is only one value of p^2 + q^2 + r^2 , we only need to consider one successful case, which, in my case, was (p, q, r) = (19, 3, 11)

p(qr) =19(p+q+r)

consider p = 19

then qr = p+q+r

since p,q,r are prime nos.

it is possible iff

when q=11,3

      r =3,11

p=19

q=3

r=11

ans =361 +121 + 9

   =491