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The Problem

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...)


Some Background

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:

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.)


My Questions

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,

  1. Does the previous step count as the base case for this step, with $n=2$? If not, what should my base case look like?
  2. What exactly should my inductive step look like here?

As always, your help is very appreciated!

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  1. Sort of, the previous step acts as both a proof for the base case (with $n=2$) and as the basis for the inductive step.

  2. You assume that "if each $m_1, \ldots, m_n$ can be written as a sum of two squares then so can their product". You want to show that this assumption implies "if each $m_1,\ldots,m_{n}, m_{n+1}$ can be written as a sum of two squares, then so can their product.".

This isn't as hard as it looks. You know that $m_1m_2\cdots m_n$can be written as a sum of squares by using your induction hypothesis. You also know that $m_{n+1}$ can be written as a sum of two squares.

So can $$\underbrace{m_1 m_2 \cdots m_n}_{\text{sum of 2 squares}} \cdot \underbrace{m_{n+1}}_{\text{sum of 2 squares}}$$ be written as a sum of two squares? (think back to the previous part now, consider $m_1\cdots m_n$ and $m_{n+1}$ as your two numbers).

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You are right. "The previous step" covers the induction case, since we are dealing with two numbers $m_1,m_2$ over there.

For induction, consider $k+1$ numbers $m_1, ...,m_{k+1}$ that can be written as the sum of two squares.

In the inductive step, you can assume that the product of $m_1, m_2, \ldots,m_k$ can be written as a sum of two squares (since here we can assume that for all integers $< k+1$, the premise is true).

Now, all you need to do, is to apply step two (or the inductive case $k=2$), with the two numbers $m_1 ... m_k$ (the product) and the last number $m_{k+1}$. That tells you that their product, which is $m_1 ... m_{k+1}$, is writable as a sum of two squares, which was to be proved.

So you were right, nothing more than step $2$ is required for the proof of the third part.

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