Prove: If each of $m_1,\dotsc,m_n$ can be written as sums of two squares, so can their product $m_1 \dotsm m_n.$ (Hint: Use induction.)
(I believe that $m_1\dots m_n$ represent composite numbers in this case...)
This problem is actually the third part of a multistep problem. The previous step had us do the following:
Prove: If $m_1$ and $m_2$ can be written as sums of two squares, so can their product $m_1 m_2$.
To which I offered the following proof:
Suppose $m_1$ and $m_2$ can be written as sums of two squares, where $m_1=a^2+b^2$ and $m_2=c^2+d^2$. Then the product $m_1m_2=(a^2+b^2)(c^2+d^2)$. Having just showed that $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$, we have $m_1m_2=(ac-bd)^2+(ad+bc)^2$. Letting $j=ac-bd$ and $k=ad+bc$, we have $m_1m_2=j^2+k^2$, which is the sum of two squares. $\square$
(The part where I used $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$ was proven in the very first step of this problem.)
I feel that this question shouldn't be much more difficult than the previous step, but I am failing to see how to apply induction here, and thus have been struggling to see how I should even start the problem off. More specifically,
- Does the previous step count as the base case for this step, with $n=2$? If not, what should my base case look like?
- What exactly should my inductive step look like here?
As always, your help is very appreciated!