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 $p^2\equiv 4$.


*

*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:)
 A: Using modular arithmetic like you do is a great way to simplify the problem.  While we eliminate the most terms by working modulo 5 as you do, we also eliminate terms working modulo 2, where the equation becomes
$$p^2+p+pq\equiv 0 \pmod 2,$$ and since $p^2+p\equiv 0 \pmod 2$, this simplifies further to $pq \equiv 0 \pmod 2$, which tells you that either $p$ or $q$ is even.  If you restrict to solutions where $p$ and $q$ are both prime, either $p=2$ or $q=2$.  In the first case, we have
$$12q=1454,$$
and in the second case, we have
$$p^2+277p=1988.$$ 
Both of these equations are straight-forward to solve.
A: Suppose that $p>7$. Then $p\ge 11$ so that $p^2+pq+275p+10q\ge 3188$ since $q\ge 2$. Hence $p\le 7$ and we have to check $p=2,3,5,7$. It follows that $(p,q)=(7,2)$.
A: This isn't the most clever approach, but it is a universal approach to quadratic diophantine equations to start by removing linear terms (completing the square).
$$
\begin{align}
p^2+pq+275p+10q
&=2008 
\end{align}
$$
Shift to remove linear terms: $p=u+a,q=v+b$
$$
\begin{align}
(u+a)^2+(u+a)(v+b)+275(u+a)+10(v+b)
&=2008\\
u^2+uv+(2a+b+275)u+(a+10)v
&=2008-a^2-ab-275a-10b
\end{align}
$$
Now we can set $a=-10$ and $b=-255$ to eliminate the linear terms:
$$
\begin{align}
u^2+uv
&=2008-100-2550+2750+2550\\
u(u+v)&=4658=2\cdot17\cdot137
\end{align}
$$
There are only $16$ possible integers solutions for $(u,v)$, which then are shifted to corresponding values for $(p,q)$. You can inspect which pairs $(p,q)$ are positive primes. Actually since you know you are looking for positive $p$, and $p=u-10$, then $u$ is at least $12$, so there are only six solutions for $u$: $17,34,137,274,2329$, and $4658$. This is just a shortcut to save yourself some checking.
A: 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.
A: $p^2+pq+275p$ is even. If $p=2$, then $q\not\in\mathbb Z$. Otherwise $p+q+275$ is even. This implies $q=2$. Then $p=7$. So the only solution is $(p,q)=(7,2)$.
