Question 618732: Find all solutions (x,y) in positive integers to:
13x + 14y = 2008
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! 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)
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