Question 618732
By Bézout's identity, we can show that there are infinitely many integer solutions (but only finitely many positive integer solutions).


One way to do it is note that (148,6) works (I wrote the equation modulo 13, showed that y = 6).


For any solution (x,y), the ordered pair (x-14, y+13) works. Since 13 and 14 are relatively prime, we can't have any other "intermediate" integer solutions. Hence, we have


(148,6)
(134,19)
(120, 32)
(106, 45)
(92, 58)
(78, 71)
(64, 84)
(50, 97)
(36, 110)
(22, 123)
(8, 136)