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Working on solving basic proofs using mathematical induction. Please see proof below. My solution took a bit of creativity (at least from my point of view). I'm hoping someone can validate this answer and provide some insights for how I may arrive at the inductive step more quickly... it took me a bit of time on scratch paper to develop a relationship between $(a^{2n+1}+b^{2n+1})$ and $(a^{2(n+1)+1}+b^{2(n+1)+1})$.

Prove that for all integers $a$ and $b$ and all $n \in \mathbb{N}, (a+b)\vert (a^{2n+1}+b^{2n+1})$.

$Proof.$ We will prove this via mathematical induction.

Base Case. Assume $a,b \in \mathbb{Z}$ and $a\neq -b$. Let $n=0$ and observe that $(a+b)\vert (a+b)$ and thus our proposition holds for $n=0$.

Inductive Step. Assume our proposition is true for $n= k$ where $k \ge 0$ and let $a,b \in \mathbb{Z}$ and $a\neq -b.$ Since $(a+b)\vert (a^{2k+1}+b^{2k+1})$ we know that $\exists c \in\mathbb{Z},(a^{2k+1}+b^{2k+1})=(a+b)c$. But then

$$\begin{align} (a^{2k+3}+b^{2k+3})&=a^2(a^{2k+1}+b^{2k+1})-b^{2k+1}(a^2-b^2)\\ &=a^2(a+b)c-b^{2k+1}(a+b)(a-b)\\ &=(a+b)(a^2c-b^{2k+1}(a-b)).\\ \end{align}$$

Thus $(a^{2k+3}+b^{2k+3})=(a+b)d$ where $d=(a^2c-b^{2k+1}(a-b))\in\mathbb{Z}$ which implies $(a+b) \vert (a^{2k+3}+b^{2k+3}$).

It follows by mathematical induction that $\forall a,b \in \mathbb{Z} \land \forall n \in \mathbb{N}, (a+b)\vert (a^{2n+1}+b^{2n+1})$. $\Box$

ClownInTheMoon
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    This checks out for me!

    I probably wouldn't have thought of setting $(a^{2k+1}+b^{2k+1})=(a+b)c$, and that drastically simplifies the argument I would have made. Additionally:

    $$a^{2k+1}+b^{2k+1}=(a+b) \left(\sum_{i=0}^{2k}(-1)^{i}a^{2k-i}b^{i}\right)$$

    If you wish for an explicit value for $c$.

     – Giuseppe Dec 15 '16 at 17:01

1 Answers1

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Yes, that works. We can make the inductive step more arithmetically intuitive by usng congruences. Namely, it follows immediately by using $\,\rm{\color{#C00}{CPR}} = $ Congruence Product Rule to multiply the first two congruences below

$$\begin{align} {\rm mod}\,\ a\!+\!b\!:\qquad\ a^2\ &\equiv\,\ b^{\large 2}\quad {\rm by}\quad \left[\,a\,\equiv\, -b\,\right]^{\large 2}\\[0.3em] a^{\large 2K+1}&\equiv\, -b^{\large 2K+1}\ \ \ {\rm i.e.}\ \ P(K)\\ \overset{\rm{\color{#C00}{CPR}}}\Longrightarrow\ \ \ a^{\large 2K+3}&\equiv\, -b^{\large 2K+3}\ \ \ {\rm i.e.}\ \ P(K\!+\!1) \end{align}$$

Your proof is essentially equivalent to that, i.e. if we take the proof of the Congruence Product Rule and specialize the values as here then it will be essentially the same as your proof. For much further discussion on this viewpoint see this answer and its links.

Community
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Bill Dubuque
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