Simultaneous Prime Equations

Relevant wiki: Quadratic Diophantine Equations - Solve by Factoring

Since we want maximum value we assume $p,q,r$ to be positive primes. Rewrite second equation as $7p+q=rp+r$ and add this to the given first equation to get $12p=rp+2r+1 \Rightarrow p(12-r)=2r+1$ . Note that since RHS and $p$ is positive we have $r<12$ so lets first check for $r=11$ substituting this in the above equations we get $p=23$ and $q=103$ . We see that as we take smaller values of $r$ , the value of $p,q$ reduces, so $(p,q,r)=(23,103,11)$ gives the maximum $p+q+r=23+103+11=\boxed{137}$

I'll call first equation A and second equation B.

Substitute the value of r from A in B to get $5p^2 -p(q+3)-(2q+1) =0$

Consider the new equation as a quadratic in $p$ .

For $p$ to be an integer, its discriminant must be a perfect square.(Let's say $k$ )

$\implies (q+3)^2 -4(5)(-(2q+1)) = k^2$

$500=(23+q+k)(23+q-k)$

Using some modular arithmetic and breaking 500 into its factors, we can get possible values of $q$ as $103$ and $7$ .

Thus after getting values of q, we can solve for p and r.

Of all triples, ${(p,q,r)}={(23,103,11)}$ gives maximum sum as $137$ .

I started basically the same as Harsh Shrivastava, solving the first equation for $r$ plugging it into the second equation, and cross-multiplying to get an equation in $p$ and $q$ .

However, when I got the quadratic in $p$ and $q$ I multiplied it out to get $5p^2-3p-2q-pq-1=0$ and then basically factored/completed the square (ignoring the constant) as $(5p-q-13)(p+2)+25=0$ . Thus, $(5p-q-13)(p+2)=-25$ .

This means that these two multiplied terms must be factors of $-25$ , but $p>0$ , so $p+2=5$ or $p+2=25$ . From this, $p=3$ or $23$ , and plugging back in gives $q=7$ or $103$ , respectively, and then $r=7$ or $11$ , respectively. The larger sum is $23+103+11=\boxed{137}$ .

7p + q = (p + 1) × r
= (p + 1)(5p – q – 1)
= 5p² + 4p – 1 – pq – q
pq + 2q = 5p² – 3p – 1
= q × (p + 2)
q = (5p² – 3p – 1) / (p + 2)
= (5p – 13) + 25/(p + 2)

25/(p + 2) must be a positive integer, and p a prime, so p = 3 or p = 23.

p = 3, q = 7, r = 7
or
p = 23, q = 103, r = 11

3 + 7 + 7 < 23 + 103 + 11

Answer = 137

i made the same steps as Nihar Mahajan but when i arrived at 12p = rp + 2r + 1 , i took r common factor instead which leads to the equation : r = 12 - $\frac{25}{p+2}$ so either p + 2 = 25 or p + 2 = 5 , by solving for q and r , we find that either p = 3 , r = 7 , q = 7 or q = 103 , r = 11 , p = 23 so the answer is 103 + 11 + 23 = 137