Let $p_1, p_2, p_3, p_4, p_5, p_6$ be consecutive primes in ascending order.
For what $p_1$ do $$p_1x+p_2y=p_3$$$$p_4x+p_5y=p_6$$ produce integer solutions $(x, y)$?
Example: $p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, p_6=13$ give $x=16, y=-9$ which are integers.
Solving gives $$x=\frac{p_3p_5-p_2p_6}{p_1p_5-p_2p_4}\quad\text{and}\quad y=\frac{p_3-p_1x}{p_2}=\frac{p_1p_6-p_3p_4}{p_1p_5-p_2p_4}$$ However, I do not know how to find when $(p_1p_5-p_2p_4)|(p_3p_5-p_2p_6)$.
By trial and error, I have found the following solutions: $$p_1=2\implies(x,y)=(16,-9)$$$$p_1=5\implies(x,y)=(-9,-8)$$$$p_1=19\implies(x,y)=(-13,12)$$$$p_1=59\implies(x,y)=(-3,4)$$ I believe there are infinitely many $p_1$ that work.