Let $k > 10$ an integer and $0 \leq x \leq k-4$ an integer and even, defined as
$$x_1= k-4 - \frac{(3k - 2)}{\sqrt{\left(4k - 3\right)}}$$
I need to prove that $x_1$ is not an integer, I have done a few tries, but it still fails. I count on your help! Thanks!
For $4k-3$ to be a square, $k$ must be odd (equal to $2n+1$ for some integer $n$), since otherwise, $4k-3\equiv5\bmod8$. So we can rewrite as $$2n-3-\frac{6n+1}{\sqrt{8n+1}}$$ This implies $n$ is a triangular number. In other words, there is some $m$ integer such that $n=0.5m(m+1)$. Substituting again, we get $$m^2+m-3-\frac{3m^2+3m+1}{2m+1}$$This can be written as $$m^2-\frac m2-\frac{15}4-\frac1{8m+4}$$
When $m=0$, $k=1$, so $m=0$ is not permitted. So, $\left\vert\frac1{8m+4}\right\vert<\frac14$. So, $-\frac72>-\frac{15}4-\frac1{8m+4}>3$, which implies that the expression is non-integral.