Is there a $k$ for which $k\cdot n\ln n$ takes only prime values? There exist some real $k$ such that $\forall $ integer $ n > 1$ the integer part of $ k *n\ln(n)$ is always prime?
 A: No.
For any choice of $k>0$ there is some $n>1$ such that $\lfloor kn\log n\rfloor$ is even (and greater than 2). Consider $kn\log n\bmod1.$
A: Here's my attempt to flesh out of Charles's answer to enhance my own confidence and understanding.  For convenience, I'm going to define $\ell(x) = x\ln x$:
First of all, note that $\ell(x+1)-\ell(x)$ $= \ln(x+1)+x\ln(x+1)-x\ln x$ $=\ln(x+1)+x\left(\ln(x+1)-\ln(x)\right)$ $= \ln(x+1)+x\ln(1+\frac1x)$.  More broadly, $\ell(x+m)-\ell(x)$ $=m\ln(x+m)+x\ln(1+\frac mx)$.  What's more, for $x\gt m$ (we'll fix $m$ later) this last quantity is bounded between $m\ln(x+m)+m-\frac{m^2}{2x}$ and $m\ln(x+m)+m$; by doing another 'iteration' of approximation, we have that $\ln x$ $\lt\ln(x+m)=\ln x+\ln(1+\frac{m}{x})$ $\lt \ln x+\frac{m}{x}$.  Piecing these together, we have that $m\ln x+m-\frac{m^2}{2x}$ $\lt \ell(x+m)-\ell(x)$ $\lt m\ln x+m+\frac{m^2}{x}$.  In particular, $\ln x+1-\frac{1}{2x}$ $\lt\left((x+1)\ln(x+1)\right)-x\ln x$ $\lt\ln x+1+\frac1x$.
Now, for a $j$ to be named later, define $\bar{j}=(-j)\bmod 1$ and $j' = \dfrac{\bar{j}}{j}$; suppose we can find an $x$ with $\frac{1}{5j}\lt (\ln x\bmod 1)-j'\lt\frac{1}{4j}$.  (It should be obvious that we can do this, as Charles notes, because of the slow growth of $\ln x$.)  Then $\bar{j}+\frac15\lt j(\ln x\bmod 1)\lt \bar{j}+\frac14$ and so $\frac15\lt j(\ln x+1)\bmod 1\lt \frac14$.  Then using the bounds above, $\frac15-\frac{j}{2x}\lt j\ell(x+1)-j\ell(x)\bmod 1\lt\frac14+\frac{j}{x}$; similarly, $\frac25-\frac{4j}{2x}\lt j\ell(x+2)-j\ell(x)\bmod 1\lt\frac12+\frac{4j}{x}$, $\frac35-\frac{9j}{2x}\lt j\ell(x+3)-j\ell(x)\bmod 1\lt\frac34+\frac{9j}{x}$, and $\frac45-\frac{16j}{2x}\lt j\ell(x+4)-j\ell(x)\bmod 1\lt1+\frac{16j}{x}$.  Choosing $x$ large enough that, e.g., $\frac{16j}{x}\lt\frac18$, then it's clear this implies that at least one of $\{j\ell(x)\bmod 1, j\ell(x+1)\bmod 1, j\ell(x+2)\bmod 1, j\ell(x+3)\bmod 1, j\ell(x+4)\bmod 1\}\in(0, \frac12)$.
But using Charles's argument, this is enough to solve the original problem: take $j=\frac k2$.  Then $j\ell(x)\bmod 1\in (0, \frac12)$ $\implies\exists m$ s.t. $m\lt j\ell(x)\lt m+\frac12$ $\implies 2m\lt k\ell(x)\lt 2m+1$ $\implies \lfloor k\ell(x)\rfloor = 2m$.  Since this is even (and it should be clear that we can take $x$ large enough that this is greater than $2$, whatever $k$ may be) then it means that $\lfloor k\ell(x)\rfloor$ can't be prime.
