$f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ 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!
 A: $\lfloor x^2 \rfloor$ - $\lfloor x \rfloor ^2$
chose one integer diferent from 1 or -1
let $ \epsilon $ be such that , you ll see why
$0<\epsilon<n-\sqrt{n^2-1}$  and
$0<\epsilon<\sqrt{n^2+1}-n$
lets show that value in this two close points differ a lot (it cannot be made small how ever we want, sorry for bad engish, i dont know correct terms )
$x_1=n+\epsilon $   ,  $x_2=n-\epsilon $
$\lfloor (n+\epsilon)^2 \rfloor = \lfloor n^2+2n\epsilon +\epsilon^2 \rfloor =n^2$ when   $2n\epsilon +\epsilon^2<1$
$\lfloor n+\epsilon \rfloor ^2 = n^2 $  when $ \epsilon<1$
so $f(x_1)=0$
now
$\lfloor (n-\epsilon)^2 \rfloor = \lfloor n^2-2n\epsilon +\epsilon^2 \rfloor =n^2-1$ when   $2n\epsilon -\epsilon^2>1$
and
$\lfloor (n-\epsilon) \rfloor^2 = (n-1)^2 $ when   $\epsilon<1$
so we get $f(x_2)=2n-2$
$\epsilon$ from start satisfies all conditions required (you get them by solving quadriatic non-equations, the ones after 'when' word) so finaly
$|f(x_1)-f(x_2)|=|2n-2|$ that means that value of function differs by a lot in some $\epsilon$ interval around integer n , that means f is discontinious in n.
i hope you understand whats the idea. :)
A: We could deal all at once with all integers $a$.  However, it is helpful to develop intuition by first looking at positive $a$.
Let  $a$ be a positive integer.   If $x$ is close enough
to $a$ but smaller than $a$, then $a^2-1\lt x^2 \lt a^2$, so $\lfloor x^2 \rfloor=a^2-1$. It follows that $f(x)=(a^2-1)-(a-1)^2=2a-2$. Thus the limit of $f(x)$ as $x$ approaches $a$ from the left is $2a-2$. Since $f(a)=0$, We conclude that $f$ is not continuous at $a$ if $2a-2\ne 0$, that is, if $a\ne 1$.  
Let $a=1$. Then the limit of $f(x)$ as $x$ approaches $a$ from the left is $2a-2$, which is $0$. It is not hard to see that the limit of $f(x)$ as $x$ approaches $a$ from the right is $0$. Also, $f(0)=0$, so $f$ is continuous at $a=0$.
Next we deal with integers $a\le 0$. Again, $f(a)=0$.  If $x$ is close enough to $a$ but larger than $a$, then $\lfloor x^2\rfloor=a^2-1$, while $(\lfloor x\rfloor)^2=(a-1)^2$. So $f(x)=2a-2$. Thus the limit of $f(x)$ as $x$ approaches $a$ from the right  is $2a-2$. However, $f(a)=0$, and $2a-2$ cannot be $0$ at $a$. 
A: 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 $\epsilon$.
