Suppose the prime numbers $p$ and $q$ satisfy $q^2+3p=197p^2+q$. Write $\frac{q}{p}$ as $l+\frac{m}{n}$, where $l,m,n$ are positive integers, $m<n$ and $\text{GCD}(m,n)=1$. Find the maximum value of $l+m+n$.

Step 1: Rearrange the equation to group prime variables:

Step 2: Since $p$ is a prime number, it must divide the left-hand side product $q(q-1)$. Since $p$ and $q$ are both prime, either $p=q$ or $p$ divides $q-1$.

• If $p=q$, then $p^2-p+3p=197p^2⟹2p=196p^2$, which has no prime solutions.

• Therefore, $p$ must divide $q-1$, so we can write:

for some positive integer $\lambda$.

Step 3: Substitute $q=\lambda p+1$ back into the equation:

Step 4: Since $p$ must be a positive prime, we must have $197-{\lambda }^2>0⟹{\lambda }^2<197⟹\lambda \le 14$.

Step 5: We test positive integers for $\lambda$ from 14 downwards:

• For $\lambda =14$:

Since 17 is a prime, $p=17$ works! Then $q=14(17)+1=239$, which is also prime.

Step 6: Calculate $\frac{q}{p}$:

Comparing to $l+\frac{m}{n}$, we get $l=14$, $m=1$, $n=17$. These satisfy the conditions $m<n$ and $\text{GCD}(1,17)=1$.

Step 7: Finally, compute $l+m+n$: