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If $a^n-b^n$ is integer for all positive integral value of n with a≠b, then a,b must also be integers.

Source: Number Theory for Mathematical Contests, Problem 201, Page 34.

Let $a=A+c$ and $b=B+d$ where A,B are integers and c,d are non-negative fractions<1.

As a-b is integer, c=d.

$a^2-b^2=(A+c)^2-(B+c)^2=A^2-B^2+2(A-B)c=I_2(say),$ where $I_2$ is an integer

So, $c=\frac{I_2-(A^2-B^2)}{2(A-B)}$ i.e., a rational fraction $=\frac{p}{q}$(say) where (p,q)=1.

When I tried to proceed for the higher values of n, things became too complex for calculation.

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Counterexample: $a=b=\pi$. – Gerry Myerson Aug 5 '12 at 9:08
may be an extra condition $a\ne b$ – pritam Aug 5 '12 at 9:16
Is there any proof for the non-trivial cases as I am rectifying the problem. – lab bhattacharjee Aug 5 '12 at 9:16

Denote $I_n=a^n-b^n$. Note that if $(a,b)$ has the property, then so has $(ma,mb)$ for $m\in\mathbb N$ as this just replaces $I_n$ with $m^nI_n$. We have $I_1\ne 0$ because $a\ne b$ and therefore find that $a=\frac{I_2+I_1^2}{2I_1}\in \mathbb Q$, say $a=\frac uv$ with $u\in\mathbb Z$, $v\in\mathbb N$. Then $w:=vb=u-vI_1$ is an integer, hence $b=\frac wv\in\mathbb Q$. If $v=1$, we are done. And if $v>1$, let $p$ be a prime dividing $v$. By the observation above about multiples, we may assume wlog. that $v=p$. Write $a=A+\frac rp$ with $0<r<p$ and $A\in \mathbb Z$. Wirth $B:=A+I_1$ we obtain $b=B+\frac rp$. From $a\ne b$ we find $A\ne B$ and hence can write $A-B=p^st$ with $s\ge 0$ and $p\not\mid t$.

Pick $m\in\mathbb N$ with $mp\ge s+3$ and $p\not\mid m$. Then $$\begin{align}(pa)^{mp}&=(pA+r)^{mp}=r^{mp}+mp^2Ar^{mp-1}+\sum_{k=2}^{mp-1}{mp\choose k}p^kA^kr^{mp-k}+p^{mp}A^{mp}\end{align}$$ and hence $$\begin{align}p^{mp}I_{mp} &=mp^2r^{mp-1}(A-B)+\sum_{k=2}^{mp-1}{mp\choose k}p^kr^{mp-k}(A^k-B^k)+p^{mp}(A^{mp}-B^{mp})\\ &=mp^{2+s}r^{mp-1}t+p^{3+s}t\sum_{k=2}^{mp-1}\frac{{mp\choose k}}pp^{k-2}r^{mp-k}\frac{A^k-B^k}{A-B}+p^{mp}(A^{mp}-B^{mp})\\\end{align}$$ where everything that is written as a fraction is in fact an integer by well-known divisibilities. Now we have a contradiction because the left hand side and all summands on the right hand side except the first are multiples of $p^{3+s}$.

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assuming $a \neq b$

if $a^n - b^n$ is integer for all $n$, then it is also integer for $n = 1$ and $n = 2$.

From there you should be able to prove that $a$ is integer.

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$a=3/2,b=1/2$, then for $n=1,2$, $a^n-b^n$ are integers – pritam Aug 5 '12 at 9:29
If $a=A+\frac{1}{2}$ where A is an integer and $b=\frac{1}{2}$, then $a^2-b^2=A^2+A$ an integer, right? – lab bhattacharjee Aug 5 '12 at 9:29
Oh, I've generalized pritam's :) – lab bhattacharjee Aug 5 '12 at 9:32
yeah, I missed the fact that I was only proving that $2*a$ was integer – molyss Aug 5 '12 at 19:17
Given three integers, $x,y,z$, $$\left(x+\frac{z}{2(x-y)}\right)^2 - \left(y+\frac{z}{2(x-y)}\right)^2$$ is an integer. – Thomas Andrews Aug 6 '14 at 22:43

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