# 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

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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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
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.
$a=3/2,b=1/2$, 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
yeah, I missed the fact that I was only proving that $2*a$ was integer –  molyss Aug 5 '12 at 19:17