# A little theorem about odd powers

Here’s an interesting little theorem I came across in number theory. By instantiating $n$, one can generate an unlimited of exercises for maths students!

Theorem. When two numbers are raised to the same odd power, the sum is divisible by the sum of the two numbers; that is: $a+b\mid a^{2n+1}+b^{2n+1}$

for any natural numbers $a$, $b$ and $n$.

Proof. We use mathematical induction on $n$. Let $\displaystyle \Phi(n) = a+b\mid a^{2n+1}+b^{2n+1}$.

The base case $\Phi(0)$ requires $a+b\mid a^1 + b^1$ – a triviality. For the inductive step, assume $\Phi(n)$ holds for arbitrary $n\ge 0$. We are required to prove $\Phi(n+1)$. First note the equivalence $a^{2(n+1)+1}+b^{2(n+1)+1} = a^2(a^{2n+1}+b^{2n+1}) + (a+b)(b-a)b^{2n+1}.$

Then observe that the first term on the RHS is divisible by $a+b$ (thanks to the induction hypothesis), and that the second term is too, since it contains $a+b$ as a factor. $\square$

Alternative proof. The above proof is concise and sufficient, but the following proof, which exploits a connection to the Binomial Theorem, is perhaps more interesting. The Binomial Theorem states: $(a+b)^n = \sum_{k\in[0,n]} {n \choose k} a^{n-k}b^k.$

We can reorder the summation so as to group the first term with the last, the second with the second-last, etc. It is helpful to consider separately the cases where the power is even and odd. Here is the case for an odd power: $(a+b)^{2n+1} = \sum_{k\in[0,n]} {2n+1 \choose n-k}(ab)^{n-k}(a^{2k+1}+b^{2k+1})$.

Now extract from the summation the $k=n$ case, and rearrange to obtain: $a^{2n+1}+b^{2n+1} = (a+b)^{2n+1} - \sum_{k\in[0,n)} {2n+1 \choose n-k}(ab)^{n-k}(a^{2k+1}+b^{2k+1})$.

We now show that $\Phi(n)$ holds for all $n$ by complete induction. Take arbitrary $n\ge 0$ and assume $\Phi(i)$ holds for all $i\in[0,n)$. We are to show that $\Phi(n)$ must hold too. $\displaystyle \begin{array}{rcll} \Phi(n) &\Leftrightarrow& a+b \mid a^{2n+1} + b^{2n+1} \\ &\Leftrightarrow& a+b \mid (a+b)^{2n+1} - \sum_{k\in[0,n)} {2n+1 \choose n-k}(ab)^{n-k}(a^{2k+1}+b^{2k+1}) \\ &\Leftarrow& a+b \mid a^{2i+1}+b^{2i+1} \text{ for all } i\in[0,n) \end{array}$

But the final statement is exactly our inductive hypothesis, so the proof is complete. $\square$