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 $$\gcd(m,n)=1$$. Find the maximum value of $$l+m+n$$.

Rewrite the given relation as a quadratic in $$q$$.

$$q^{2}+3p=197p^{2}+q \;\Longrightarrow\; q^{2}-q+\bigl(3p-197p^{2}\bigr)=0 \; -(1)$$

For integer (in fact, prime) $$q$$, the discriminant of $$-(1)$$ must be a perfect square.

Discriminant $$D = 1-4\bigl(3p-197p^{2}\bigr)=788p^{2}-12p+1 = k^{2}$$ for some integer $$k$$.
Hence

$$k^{2}-788p^{2}=1-12p \; -(2)$$

Equation $$-(2)$$ shows $$k$$ is very close to $$\sqrt{788}\,p \approx 28.089p$$. Write $$k=28p+t$$ where $$t$$ is an integer expected to be small.

Substituting $$k=28p+t$$ into $$-(2)$$ gives

$$(28p+t)^{2}-788p^{2}=1-12p$$ $$\Longrightarrow 784p^{2}+56tp+t^{2}-788p^{2}=1-12p$$ $$\Longrightarrow 4p^{2}-(56t+12)p+(1-t^{2})=0 \; -(3)$$

For integer $$p$$, the discriminant of $$-(3)$$ must also be a perfect square:

$$\Delta = (56t+12)^{2}-4\!\cdot\!4(1-t^{2}) = 16\!\bigl(197t^{2}+84t+8\bigr)=16s^{2}$$ $$\Longrightarrow 197t^{2}+84t+8=s^{2} \; -(4)$$

Check small integers $$t$$ in $$-(4)$$:

• $$t=0:\; s^{2}=8$$ (not a square)
• $$t=1:\; s^{2}=197+84+8=289=17^{2}$$ ✔️
• $$t=-1:\; s^{2}=197-84+8=121=11^{2}$$ ✔️

Insert each valid $$t$$ into $$-(3)$$.

$$p=\dfrac{56t+12\pm4s}{8}=\dfrac{56(1)+12\pm68}{8}=\dfrac{68\pm68}{8}$$ Gives $$p=17$$ (positive) or $$p=0$$ (rejected). Thus $$p=17$$ (prime) and $$k=28p+t=28\!\cdot\!17+1=477$$. Using $$q=\dfrac{1+k}{2}$$, obtain $$q=\dfrac{478}{2}=239$$ (prime).

$$p=\dfrac{-56+12\pm44}{8}=\dfrac{-44\pm44}{8}$$ gives $$p=0$$ or $$p=-11$$, both rejected.

Therefore the only admissible prime pair is $$p=17,\; q=239$$.

The required ratio is

$$\frac{q}{p}=\frac{239}{17}=14+\frac{1}{17}$$

Here $$l=14,\; m=1,\; n=17$$, so

$$l+m+n = 14+1+17 = 32$$

As the derivation shows there is no other prime solution, this $$32$$ is the maximum (and only) possible value.

Final Answer: 32