$f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at x=1 How to prove that $f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at $x=1$ ?
Ya,even I used intuition at the first go taking some trial values and checking.Then I plotted the graph on wolfram alpha.But if you can think of some other rigorous proof for this one,let me know.Thanks!

The approach here is to find formula that are equivalent to $f$ locally. The idea being that it is simpler to spot the discontinuity from the locally
equivalent formula.
Note that $\lfloor x \rfloor = 1_{[0,\infty)}(x)-1$ for $|x|<1$. In particular,
for small positive $x$, we have $\lfloor x \rfloor =0$. Also note that
$\lfloor x \rfloor^2 = - \lfloor x \rfloor$ for small $x$.
Consider $f(n+\epsilon)$ for integer $n$ and small $\epsilon$.
If $n=0$, we have $\lfloor \epsilon^2 \rfloor = 0$.
Hence
$f(\epsilon) = -\lfloor \epsilon \rfloor^2 = \lfloor \epsilon \rfloor$ for small $\epsilon$, and so $f$ is discontinuous at $x=0$.
For $n \neq 0$ and sufficiently small $\epsilon$,
we have $\lfloor (n+\epsilon)^2 \rfloor = n^2 +\lfloor n \epsilon + \epsilon^2 \rfloor= n^2 + \lfloor n \epsilon\rfloor$, and
$\lfloor n+\epsilon\rfloor = n + \lfloor \epsilon\rfloor$. Hence
$\lfloor n+\epsilon\rfloor^2 = n^2 + 2n \lfloor \epsilon\rfloor + \lfloor \epsilon\rfloor^2$,
and so
$f(n+\epsilon) = \lfloor n \epsilon\rfloor -2n \lfloor \epsilon\rfloor - \lfloor \epsilon\rfloor^2 = \lfloor n \epsilon\rfloor -(2n-1) \lfloor \epsilon\rfloor$.
If $n<0$ this gives
$f(n+\epsilon) =\lfloor- \epsilon\rfloor +(1-2n) \lfloor \epsilon\rfloor$, from which we see that $f(n) = 0$ and $f(n+\epsilon) =-1$ for small positive $\epsilon$.
If $n>0$, we have $f(n+\epsilon) =\lfloor \epsilon\rfloor +(1-2n) \lfloor \epsilon\rfloor = 2(1-n) \lfloor \epsilon\rfloor$, hence $f$ is continuous
for $n = 1$, otherwise $f(n) = 0$ and $f(n+\epsilon) = -2(n-1)$ for small
negative $\epsilon$.
