Let $ x^{2}-x = m\in\mathbb{Z}$. Then $ x =\frac{1\pm\sqrt D}{2}$, where $ D = 4m+1 > 0$. If $ \sqrt{D}$ isn't irrational, clearly results $ x\in\mathbb{Z}$. Assume the contrary. Take firstly $ x =\frac{1+\sqrt D}{2}$. Then $ x^{n}-x =\left(\frac{1+\sqrt D}{2}\right)^{n}-\frac{1+\sqrt D}{2}=\frac{1}{2^{n}}\left[(1+\sqrt D)^{n}-2^{n-1}(1+\sqrt D)\right]$ is an integer. This implies the weaker assertion that $ (1+\sqrt D)^{n}-2^{n-1}\sqrt D$ is an integer. Because $ (\sqrt D)^{2}\in\mathbb{Z}$, we must consider only the odd exponents in the expansion of $ (1+\sqrt D)^{n}$. So, we have that $ \sqrt D\left[-2^{n-1}+\sum_{k = 0}^{\left[\frac{n-1}{2}\right]}D^{k}\binom{n}{2k+1}\right]$ is an integer. It results that $ E =-2^{n-1}+\sum_{k = 0}^{\left[\frac{n-1}{2}\right]}D^{k}\binom{n}{2k+1}$ is $ 0$. However, we have $ D\ge4\cdot1+1 = 5$, so $ E >\sum_{k = 0}^{\left[\frac{n-1}{2}\right]}\binom{n}{2k+1}-2^{n-1}= 0$. Contradiction. The case $ x =\frac{1-\sqrt D}{2}$ is treated completely analogously. Because when taking odd exponents, all the binomial coefficients will have a $ -$ in front of them, and the 'separate' $ \sqrt D$ we'll have the coefficient $ 2^{n-1}$, not $ -2^{n-1}$ we'll obtain a similar contradiction getting this time $ E < 0$. Just as a commentary, I think this problem belongs more to the $ F$ section - Rational and irrational numbers.

Let $ r_1 > r_2$ be the solutions to the quadratic equation $ y^2-y=\lambda$, where: $ \lambda=x^2-x$ (the minimum of $ y^2-y$ is -0.25, so $ \lambda$ is non-negative). Induction shows that $ x^n=a_{n}x+b_{n}$ for all $ n \in N$, where: $ a_{n+2}=a_{n+1}+\lambda a_{n}$, $ b_{n+2}=b_{n+1}+\lambda b_{n}, a_{0}=0, a_{1}=1, b_{0}=1, b_{1}=0$ (an important thing is that they also satisfy $ a_{n+1}=b_{n}+a_{n}, b_{n+1}=\lambda a_{n}$) Notice that $ a_{n}, b_{n}$ are integer sequences. So, if $ x^n-x$ is an integer, it implies that either x is a rational number (because $ (a_{n}-1)x+b_{n}, a_{n}, b_{n} \in Z$), or $ a_{n}=1$. First case: x is rational. $ x^2-x-\lambda=0 \implies \frac{1\pm\sqrt{1+4\lambda}}{2} \in Q \implies 1+4\lambda$ is a square, or: x is an integer (because this square is a odd square and its root is also odd). Second case: $ a_{n}=1$. From the recursion $ a_{n+1}=b_{n}+a_{n}, b_{n+1}=\lambda a_{n}$, we have that $ a_{n}, b_{n}$ are positive integers for n>2, and that $ a_{n+1}=b_{n}+a_{n}>a_{n} \implies a_{n} \ge a_{3}=1+\lambda$ for $ n \ge 3$. If $ \lambda=0$, we have $ x^2-x=0$, or: $ x=0,1 \in Z$. Else, $ a_{n}>1$ and we get a contradiction.