<s>[INST] Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rewrite also as: $p^2+pq\equiv 3 \pmod5$.
Now I analyzed five possible cases: $p^2\equiv 0$, $p^2\equiv 1$, $p^2\equiv 2$, $p^2\equiv 3$ and $p^2\equiv 4$.


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*We can not that if $p^2\equiv 0$ also $pq\equiv 0$ therefore the equation is impossible

*$p^2\equiv 2$ is impossible because $2$ is a  non-quadratic residue $\pmod5$

*also $p^2\equiv 3$ is impossible because $3$ is a non-quadratic residue $\pmod5$.

*If $p^2\equiv 1 \pmod5$, $p\equiv \pm 1$, therefore $p\equiv -1\equiv 4 \pmod5$ and  $p2\equiv 1 \pmod5$. These values are not acceptable because if $p=9$ ($9\equiv 4 \pmod5$) $275\cdot 9>2008$, if $p=11$ ($11\equiv 1\pmod5$) $275\cdot 9>2008$ therefore two values aren't acceptable.

*If $p^2\equiv 4 \pmod5$, $p\equiv \pm2$ and if $p\equiv -2\equiv 3 \pmod5$ there not exist solutions for the same reason of previous example. If $p\equiv 2 \pmod5$ possible values of $p$ are $2$ or $7$ because $275p<2008$. Replacing $p=2$ we obtain a valor not integer of $q$ while replacing $p=7$ we obtain $q=2$ therefore this is the only solution of equation. 


Is correct my process or are there other solutions?
Thanks:)
 [/INST] Because there is alot of solutions, let me present you mine: the equation you stated is equivalent to $$(p+10)(p+q+265)=4658=2\cdot 17\cdot 137$$
and because $p+10>2$ then $p+10\geq 10$ but also $p+q+265\geq 2\cdot 137$ hence $p+2=17$ and $p+q+265=2\cdot 137$

In negative solutions (without (negative) prime assumptions) there is exactly $16$ solutions, because every factor can take any possible value of the divisors of $4658$ and there are exactly $16$ possible divisors.
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