TITLE: Solve $\lceil x\rceil^2+\lceil x+1\rceil^2=25$
QUESTION [5 upvotes]: How can we find the solution set of $\lceil x\rceil^2+\lceil x+1\rceil^2=25$ where $\lceil x\rceil$ is the ceiling function?

REPLY [8 votes]: Observe that $\lceil x+1\rceil =\lceil x\rceil +1$. So setting $n=\lceil x\rceil$, it amounts to solving for $n$ in $$n^2+(n+1)^2=25\;\iff\;n^2+n-12=0\;\iff\;(n+4)(n-3)=0\;\iff\; n\in\{-4,3\}.$$
The case $n=\lceil x\rceil=-4$ gives you $-5<x\leq-4$ while $n=\lceil x\rceil=3$ yields $2<x\leq 3$. So the solution set is
$$ (-5,-4]\cup(2,3].$$