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

Question

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.

Answer 1

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.

Answer 2

Here's a sketch of the answer. If there is a counterexample, it has the form $a=A+q+ri$ and $b=q+ri$, where $A$ is an integer, $q,r$ rationals, and $i=\sqrt{-1}$. The $n=2$ case quickly rules out a complex example, so $r=0$. Finally, each $n>2$ shows that $A$ would have to be an arbitrarily large power of the denominator of $q$, contradiction.

EDIT: As noted below, this answer has problems. I'm marking myself wrong until I can rethink it.