About a proof that $\lfloor x^2\rfloor = \lfloor x\rfloor^2$ for unbounded non integer values of $x$ I am taking a first course in discrete mathematics. The instructor parsed the following question that has the following solution, respectively:
Prove the statement:

For all positive integers $N$, there exists a real number $x$, so that $x$ is not an integer, $x>N$, and $\lfloor x^2\rfloor = \lfloor x\rfloor^2$.


Why is $x= N+ 1/(3N)$ used? I do not understand this statement. Is $1/(3N)$ some axiom of non-integer real numbers?
Also, what is the intuition behind squaring both sides at the end of the third line of the solution?
 A: If you feel like numbers and expressions are appearing out of nowhere for no apparent reason,  it's because they are. It's part of the magic show of a completed proof to get a round of applause at the end when all the various pieces come together and fall neatly into place to produce the desired answer.
What the proof could have done to reduce the suspense but perhaps improve understanding is to tell you at the outset that the intention is to find a real $x$ for which $\lfloor x \rfloor =N$ and $\lfloor x^2\rfloor = N^2$, by making $x$ just a tiny fraction larger than $N$. The exploratory part - seeing just how tiny that fraction has to be - is left out of the proof; you just get to see the fruits of that investigation, a tiny but well-defined fraction that does the job.
A: Since we want $x>N$, we can set $x=N+y$ with $y>0$. We also want
$$
\lfloor (N+y)^2\rfloor=\lfloor N+y\rfloor^2
$$
and, if we take $y<1$, the right-hand side is $N^2$. Thus we need
$$
(N+y)^2<N^2+1
$$
so that also the left-hand side is $N^2$. The inequality becomes
$$
y^2+2Ny-1<0
$$
that is satisfied by
$$
-N-\sqrt{N^2+1}<y<-N+\sqrt{N^2+1}
$$
It's not difficult to show that $\sqrt{N^2+1}-N<1$, so adding the above conditions $0<y<1$, we obtain the full set of solutions:
$$
0<y<\sqrt{N^2+1}-N
$$

The book chooses another way. The authors looked for a solution of the form $y=\frac{1}{aN}$, so they wanted
$$
N^2+\frac{2}{aN}+\frac{1}{a^2N^2}<N^2+1
$$
and this becomes
$$
\frac{2}{aN}+\frac{1}{a^2N^2}<1
$$
that can be rewritten as
$$
a^2N^2-2aN-1>0
$$
This is certainly satisfied for 
$$
a>\frac{1+\sqrt{2}}{N}
$$
(just solve the quadratic). The maximum value of the bound for $a$ is when $N=1$, that is, $1+\sqrt{2}$ which is less than $3$. Then they could take $a=3$ and do the steps backwards in order to surprise the readers like when magicians pull a rabbit out of a top hat.
