Any ideas?

gilbert wrote: Any ideas? My solution at http://www.math.vn/showthread.php?p=53292&posted=1#post53292 and http://vn.360plus.yahoo.com/math-2M/article?mid=40

Any solutions?

An idea is obviously induction! We get easily that $x_n\in Q$ and \[ x_n\equiv \frac{2(a^2+b^2)^k}{(a+b)(2(a^2+b^2)^{k-1})} \] for some $k\in N$ and for any $n\in N$. Suppose that for $m\in N$ and $x_m$ is integer, it's equivalent to $ \frac{2(a^2+b^2)^k}{a+b} \in Z $ and by $gcd(a,b)=1 $, $a,b>1$ we get a contradiction.

Thank you mathuz:-)

Any other solution?

anyone solve this