We will consider as follows:Case y

We will consider as follows:
Case y = 0 : We have 19x +1 = z2 . By Lemma 2.2, the Diophantine equation
19x +1= z2 has no non-negative integer solution.
Case y = 1: We have 19x + 2 = z2 . Let us consider z as odd and even
numbers, it obvious that there is no solution.
Case y >1: If x = 0 , we have 1+ 2y = z2 . By lemma 2.3, the solution (3,3) is a
unique solution ( y, z) for the Diophantine equation z2 − 2y =1.
Hence, x ≥1. Note that z is odd. Then z2 ≡1(mod4). Since y ≥ 2 ,
so 2y ≡ 0 (mod 4) . Thus 19x ≡1 (mod 4) . Since 19 ≡ 3 (mod 4)
and 192 ≡ 33 ≡1 (mod 4) , so , but 192n+1 ≡ 3 (mod4).
Hence, x is even. Let x = 2k where k is a positive integer.
Then z2 −192k = 2y . Then (z −19k )(z +19k )= 2y . So z −19k = 2u
and z +19k = 2y−u where u is a non-negative integer. Then
2(19k )= 2y−u − 2u = 2u (2y−2u −1) where y > 2u . It follows that u =1.
Then 19k = 2y−2u −1. That is 2y−2u −19k =1. Since y > 2u , so
y − 2u ≥1. If y − 2u =1, we have 2 −19k = 1 or 19k = 1 which
contradicts with k is a positive integer. Thus y − 2u >1. Hence, by
Proposition 2.1, the Diophantine equation 2y−2u −19k =1 where
y − 2u > 1 and k >1 has no non-negative integer solution. Thus in
case k =1, we have 2y−2u = 20 which is impossible.
Therefore, the solution (0,3,3) is a unique solution (x, y, z) for the
Diophantine equation 19x + 2y = z2 where x, y and z are non-negative integer.