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.
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$\begingroup$ Hint: Let $t=(x+y)$ $\endgroup$– nonuserMay 1, 2021 at 15:57
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$\begingroup$ There are infinite many solutions. You can see this by putting $x = -y$ so that the equation reduces to $2x + 1 = z^2$. This one requires that $z$ be odd so putting $z=2k+1$ one gets $x=2k^2+k$ and $y=-2k^2-k$ where k is any integer. $\endgroup$– SalcioMay 1, 2021 at 16:06
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$\begingroup$ You can not do that since $x,y$ are natural. @Salcio $\endgroup$– nonuserMay 1, 2021 at 16:07
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$\begingroup$ OK, then $k$ is a natural number ... $\endgroup$– SalcioMay 1, 2021 at 16:08
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$\begingroup$ ???????? $$x=X-y$$ $$X^2+3X-2y=z^2$$ $$y=\frac{1}{2}(X^2-z^2)+X$$ $\endgroup$– individMay 7, 2021 at 18:04
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2 Answers
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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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Yes, inequalities should be implemented. Let $t=x+y$ then we have $$z^2 = t^2+t+1+2x$$
Notice that $0<x<t$ so we have $$t^2< t^2+t+1+2x <t^2+4t+4$$ which means $$t^2<z^2<(t+2)^2\implies z=t+1$$ Putting this in OP we get $$t =2x\implies x=y$$
So we have infinte triples, for each $x\in \mathbb{N}$ we have $(x,x,2x+1)$