finding the integral solutions

Question

Prove that the equation $\sum\limits_{i=1}^n x_i^{-2} = 1$ has integral solutions for $n > 6$.

I have no idea how to proceed, can someone give me a hint.

Sorry for the mistakes in the text above.

Answer 1

For $n=7$, take $x_1=x_2=x_3=x_4=4$ and $x_5=x_6=x_7=2$

For $n=8$, take $x_1=x_2=2$, $x_3=x_4=x_5=x_6=3$ and $x_7=x_8=6$

For $n=9$, take $x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=x_9=3$

Now suppose we have a solution $x_1, x_2$ . . . $x_n$ for some $n$. Then:

$\sum\limits_{i=1}^nx_i^{-2}=1$

So, $\sum\limits_{i=1}^n(2x_i)^{-2}=1/4$.

Therefore, $2x_1, 2x_2$ . . . $2x_n, 2, 2, 2$ becomes a solution for case of $n+3$. As the problem has been solved for $n=7,8,9$, by induction, we have a solution for all $n>6$.