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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.

  • 3
    Counterexample: $a=b=\pi$.2012-08-05
  • 2
    may be an extra condition $a\ne b$2012-08-05
  • 0
    Is there any proof for the non-trivial cases as I am rectifying the problem.2012-08-05
  • 3
    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=488&t=1508432012-08-05

2 Answers 2