solutions to linear equations involving prime numbers? Suppose we have the two equations:

$2Z - p = Xq$
$2Z - q = Yp$

where $X,Y,Z \in \mathbb{N} $ and $p,q \in \mathbb{P} - \left\{2\right\} $
Are there any solutions where $Z$ isn't prime?
 A: Yes, there are infinitely many solutions with $Z$ non-prime.
One infinite family is given by considering any odd primes $p$ and $q$ such that $$p\equiv (+1)\pmod 6 \\ q\equiv (-1)\pmod 6$$ Then, $p+q\equiv 0\pmod 6$ and setting $X=Y=1$ and $Z=(p+q)/2$ yields a solution with non-prime $Z$.
If you prefer less trivial solutions (e.g. with $X$ distinct from $Y$), there are enough such too -- one example could be $(p,q,X,Y,Z)=(3,5,34,13,21)$ (useless curio: Fibonacci numbers!)
A: $$ Z \equiv p/2 \pmod q,$$
$$ Z \equiv q/2 \pmod p,$$
combine with Chinese Remainder Theorem. If you dislike $Z,$ keep adding $pq$ until you get something you like.
If this makes you uncomfortable, rewrite as
$$ Z \equiv (p+q)/2 \pmod q,$$
$$ Z \equiv (p+q)/2 \pmod p.$$
In any case, end with $X = (2Z-p) /q $ and $Y = (2Z-q) /p. $  
Now that I think about it, just take
$$ Z = \left( \frac{p+q}{2 } \right) + n p q  $$ for your favorite $n \geq 1,$ then
 $X = (2Z-p) /q $ and $Y = (2Z-q) /p. $ 
$$ p=3, q=5, X=1, Y=1, Z=4,      $$
$$ p=3, q=5, X=7, Y=11, Z=19,      $$
$$ p=3, q=5, X=13, Y=21, Z=34,      $$
$$ p=3, q=5, X=19, Y=31, Z=49,      $$
$$ p=3, q=5, X=25, Y=41, Z=64,      $$
and just keep increasing $X$ by $6,$  increasing $Y$ by $10,$  increasing $Z$ by $15.$ 
A: You get trivial counterexamples by taking $X=Y=1$ so that $Z=(p+q)/2$, which is non-prime for lots of $p$'s and $q$'s, but let's look beyond that.
It's clear that $X$ and $Y$ must be odd, so let's write $X=2m+1$ and $Y=2n+1$ with integers $m$ and $n$.  The equations now are
$$2Z=p+(2m+1)q=p+q+2mq$$
and
$$2Z=q+(2n+1)p=p+q+2np$$
from which follows that
$$mq=np$$
which implies $m=kp$ and $n=kq$, and this gives
$$Z={p+q+2kpq\over2}={p+q\over2}+kpq$$
Now let $p=3$ and $q=5$, so that $Z=4+15k$, and recall that $X=2kp+1=6k+1$ and $Y=2kq+1=10k+1$.  You get a counterexample at $k=2$.
In general, whatever you take $p$ and $q$ to be, you'll certainly have a counterexample with
$$Z=\left({p+q\over2}\right)(1+pq)$$
A: Assume $p\ne q$. Note that every integer $Z\equiv(p+q)/2\pmod{pq}$ is a solution: writing $Z=(p+q)/2+kpq$, we can set $X=2kp+1$ and $Y=2kq+1$. (This sufficient condition turns out to be necessary as well.) And every residue class contains infinitely many non-primes (in fact, most elements are non-primes). Indeed, for any integer $n$ not divisible by $p$ or $q$, there are infinitely many multiples of $n$ in the residue class $(p+q)/2\pmod{pq}$ by the Chinese remainder theorem. More interestingly, there actually are infinitely many prime values of $Z$ to be found as well (since $(p+q)/2$ and $pq$ are relatively prime).
When $p=q$, it's easy to see that $Z$ can be any multiple of $p$ and nothing else - so again, infinitely many non-primes.
