Prove that for all $n\in\mathbb{N}$, there exists a set $A_{n}=\{ (a-b),(a^2-b^2),...,(a^n-b^n)\}\colon a,b\in(\mathbb{R}\setminus\mathbb{Z})\}.$ Prove that for all $n\in\mathbb{N}$, there exists a set  $A_{n}$ of positive integers such that $$A_{n}=\{ (a-b),(a^2-b^2),...,(a^n-b^n)\ \colon\ a,b\in(\mathbb{R}\setminus\mathbb{Z})\}.$$
If we can find two real numbers $a$ and $b$ such that $$a,b\in(\mathbb{R}\setminus\mathbb{Z})\implies(a-b),(a+b),(a\cdot b)\in \mathbb{Z}$$
holds, then we'll be on our way to the solution.
 A: You need $a$ and $b$ to be rational. Thus you want the first $n$ powers
of $a$ to have the same fractional parts as the first $n$ powers of $b$.
Let's try $b=1/m$ and $a=r+1/m$ for positive integers $r$ and $m$.
Then if $m^{n-1}\mid r$ we have, by the binomial theorem, $a^n=r^n+\cdots
+n r/m^{n-1}+1/m^n$. All terms, save the last are integers. So set, say,
$a=2^{n-1}+1/2$ and $b=1/2$.
A: If $a^2-b^2$ and $a-b$ are positive integers, then $a+b=\frac{a^2-b^2}{a-b}$ must be a positive rational. It follows that $a,b$ are also rational (if $n\ge2$). Let $m$ be their common denominator, so $a=\frac cm$, $b=\frac dm$ with coprime positive integers $c,d$.
Note that fo4 $1\le k<n$,
$$\tag1a^{k+1}-b^{k+1}=(a^k-b^k)a+(a-b)b^k$$
As the first summand has denominator $\le m$, we see that $a-b$ must be divisible by a high power of $m$. In act, if $m^{n-1}\mid a-b$, we see from $(1)$ by induction that $m^{n-k}\mid a^k-b^k$ and therefore all $a^k-b^k$, $1\le k\le n$ are integers (and of course positive, provided $a>b$). So let $b$ be an arbitrary non-integer rations $b=\frac cm$ and let $a=b+m^{n-1}$.
