If $a$ can be written as sum of squares of three rational numbers, prove that $a^m$ can be too. Let a be a rational number that can be written as sum of squares of three rational numbers. Prove that $a^m$ can also can be written as sum of squares of three rational numbers for any positive integer $m$.
I tried to solve using Mathematical Induction
Let a be a rational number that can be written as sum of squares of three rational numbers
i.e. $$a=\left( p_1 / q_1\right)^2+\left( p_2 / q_2\right)^2+ \left( p_3 / q_3\right)^2\\$$
We prove $a^m$ is rational for every positive integer$\\$
For $n=1$, it is true. $\\$
Let it be true for $n=k$
i.e. $$a^m=\left( r_1 / s_1\right)^2+\left( r_2 / s_2\right)^2+ \left( r_3 / s_3\right)^2\\$$
Consider 
$$a^{m+1}=a^m *a$$
$$=(\left( p_1 / q_1\right)^2+\left( p_2 / q_2\right)^2+ \left( p_3 / q_3\right)^2) (\left( r_1 / s_1\right)^2+\left( r_2 / s_2\right)^2+ \left( r_3 / s_3\right)^2)\\$$
I am not sure how to proceed next to show that $a^{m+1}$ can be written as sum of squares of three numbers.
Thanks in advance for any kind help.
 A: Based on  (When is a rational number a sum of three squares?), a  rational number  $p/q$ is sum of three rational 
squares if and only if  $pq$ is not of the form $2^{2a}(8b-1)$.
Note that any integer $n$ can be factored as $n=2^r\cdot s$, where $r\geq 0$ and $s$ is odd.
So the result above is telling you that  if  a number $p/q$  is NOT a sum of three rational squares, then $pq=2^r\cdot s$, where    $r$  is  even and $s\equiv -1 \mod 8$.
Equivalently, if  a number $a=p/q$  is a sum of three rational squares, then  $pq=2^r\cdot s$, where either  $r$  is odd or $s\equiv 1, \pm 3 \mod 8$.
So things pretty much reduce to compare the behavior of  $s$ and $s^m$ $\mod 8$.
Suppose $a=p/q$ i sa sum of three rational squares. That is, $pq=2^rs$, where either $s\equiv 1$ or $\pm 3 \mod 8$, or $r$ is odd. In the former case, the powers $s^m$ will never be $\equiv -1 \mod 8$.
In the latter case, $s^m \equiv -1 \mod 8$ only if $m$ is odd. But  then $(pq)^m=2^{rm}\cdot s^m$, with $rm$  odd. 
Therefore, in either case, $(pq)^m$ will not be of the form  $2^ab$, where 
$a$ is even and $b\equiv -1 \mod 8$. Hence $a^m$ is the sum of three rational squares.
A: There is a simple proof requiring no infection. If $m=2k$, $a^m = (a^k)^2+0^2+0^2$; if $m=2k+1$, $a^m=(a^k)^2a=(a^kr_1)^2+(a^kr_2)^2+(a^kr_3)^2$. 
