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

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
None of the solution is acceptable, right? What to do as they are hampering the accept rate. – lab bhattacharjee Aug 12 '12 at 11:48

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

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.

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I cannot see how one would derive that $a,b\in\mathbf Q[i]$ without actually showing that $a,b\in\mathbf Q$, because $a-b$ and $a+b=\frac{a^2-b^2}{a-b}$ are rational. – Marc van Leeuwen Aug 5 '12 at 13:05
@Marc van Leeuwen: So you agree that "The $n=2$ case quickly rules out a complex example"! ;-) – trb456 Aug 5 '12 at 13:26
@trb456, could you please elaborate a little more. – lab bhattacharjee Aug 5 '12 at 15:31
@trb456: My difficulty is that you seem to implicitly ("it has the form") rule out the "complex irrational" case (i.e., in $\mathbf C\setminus\mathbf Q[i]$) before you even consider $n=2$, and I can see no reason for that. – Marc van Leeuwen Aug 5 '12 at 16:09
@Marc van Leeuwen: Brain freeze on my part, of course you are correct. I'm going to mark my post as wrong until I can rethink it. Thanks! – trb456 Aug 5 '12 at 18:16