Freeman and Tukey interval If $Y$ is Poisson$(\lambda)$ distributed, how do we obtain this confidence interval for $\lambda$:
$$
\left (\frac{(\sqrt Y+\sqrt{Y + 1}-z_{(α⁄2)} )^2 - 1}{4},\  \frac{(\sqrt Y+\sqrt{Y  +1}+z_{(α⁄2)} )^2 - 1}{4} \right )
$$
 A: Taking these as given:
$E(\sqrt{Y}+\sqrt{Y+1})\approx \sqrt{4\lambda+1}$
$\text{Var}(\sqrt{Y}+\sqrt{Y+1})\approx 1$
and assuming $\lambda$ is large enough that $(\sqrt{Y}+\sqrt{Y+1})$ is approximately normal, then
$P(-z_{1-\alpha/2}<\sqrt{4\lambda+1}-(\sqrt{Y}+\sqrt{Y+1})<z_{1-\alpha/2})\approx 1-\alpha$
$P((\sqrt{Y}+\sqrt{Y+1})-z_{1-\alpha/2}<\sqrt{4\lambda+1}<(\sqrt{Y}+\sqrt{Y+1})+z_{1-\alpha/2})\approx 1-\alpha$
Further assume that $\lambda$ and $\alpha$ are large enough that the lower limit ($(\sqrt{Y}+\sqrt{Y+1})-z_{1-\alpha/2}$) is non-negative. For $\alpha=0.05$ this means $Y$ should be at least $1$ (though for a reasonably good approximation you'll want it larger than that anyway) -- and the case $Y=0$ would require a different approximation, as it does for the double root residual. If your observed $Y$ is $0$ you simply shouldn't be using this sort of approximation there (and down that far any hope of a two-tailed $1-\alpha$ interval with $\alpha$ at typical values is gone as well).
Then we can square all the terms without complication (we won't be flipping any inequalities, having to consider separate cases for $\pm$, or introducing any extraneous solutions):
$P((\sqrt{Y}+\sqrt{Y+1}-z_{1-\alpha/2})^2<{4\lambda+1}<(\sqrt{Y}+\sqrt{Y+1}+z_{1-\alpha/2})^2)\approx 1-\alpha$
$P(\frac14 [(\sqrt{Y}+\sqrt{Y+1}-z_{1-\alpha/2})^2-1]<{\lambda}<\frac14[(\sqrt{Y}+\sqrt{Y+1}+z_{1-\alpha/2})^2-1]\approx 1-\alpha$
which is the required interval.
Note that since $\lambda$ must be positive, we now require $Y$ to be at least 2 to get even a feasible answer for the lower limit at the 5% level (though it won't give a very good approximation for the coverage down that far) --  don't try to use this at all when $Y\leq 1$. 
Indeed I'd suggest using it when $Y\leq 4$ or so isn't a great idea; once you get much above $4$ it tends to work at least reasonably well at typical levels and nearby $\lambda$ values. (Your own utility function or circumstances may vary, don't take that value as in any way prescriptive or even worth quoting).
