We now present our results.Theorem

We now present our results.
Theorem 3.1: The Diophantine equation 3x + 19y = z2
has exactly two solutions (x; y; z) in non-negative integers, i.e.
(x; y; z) ∈ {(1; 0; 2); (4; 1; 10)}:
Proof: Let x; y; z be non-negative integers. We first
consider the case when one of the unknown x; y; z is zero.
By Theorem 2.1 and Theorem 2.2, we see immediately that
(x; y; z) = (1; 0; 2): On the other hand, for x; y; z > 0 we
divide y into two cases.
Case 1. If y is even, say y = 2n for some non-negative
integer n, then (z + 19n)(z − 19n) = z2 − 192n = 3x: This
implies that 2 · 19n = (z +19n)−(z−19n) = 3 −3 where
+ = x such that  > : Hence,  = 0 and 2·19n = 3x−1:
Adding both sides by −2, we obtain 2(19n−1) = 3(3x