<s>[INST] Solution of the Diophantine equation 
What are the possible triples (x,y,z) in positive integers such >that,
$$(x+y)^{2}+3x+y+1=z^{2}$$

I have used the inequality approach and many others but wasn't able to find an answer.
 [/INST] Let $t:=x+y$ so that the Diophantine equation is equivalent to
$$t^2+t+2x+1=z^2.$$
In particular this shows that $z$ is odd, say $z=2s+1$. Multiplying by $4$ yields
$$(4s+2)^2=4z^2=4t^2+4t+8x+4=(2t+1)^2+8x+3,$$
and a bit of rearranging then shows that
$$(4s+2)^2-(2t+1)^2=8x+3.$$
Conversely, for any pair of integers $s$ and $t$ we have
$$(4s+2)^2-(2t+1)^2\equiv3\pmod{8},$$
which shows that the integral solutions are parametrized by
\begin{eqnarray*}
(x,y,z)
&=&\left(\frac{(4s+2)^2-(2t+1)^2-3}{8},t-\frac{(4s+2)^2-(2t+1)^2-3}{8},2s+1\right)\\
&=&\left(2s^2+2s-\frac{t^2+t}{2},-2s^2-2s+\frac{t^2-t}{2},2s+1\right).
\end{eqnarray*}
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