<s>[INST] $f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ 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!
 [/INST] 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 $a$. 
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