# Do I need to use induction to have sufficient rigor in this proof?

I'm taking my first analysis class this summer. The professor asked us to prove that $a^{2n}-b^{2n}$ is divisible by $a+b$. After dorking around with the first couple of $n$ I was able to come up with a general term for the factorization of $a^{2n}-b^{2n}$. What I got was:

$$a^{2n}-b^{2n}=(a+b)\sum_{k=0}^{2n-1}(-1)^ka^{2n-1-k}b^k$$ I proceeded with the proof as follows:

$$\implies \sum_0^{2n-1}(-1)^ka^{2n-k}b^k+\sum_0^{2n-1}(-1)^ka^{2n-1-k}b^{k+1}$$

$$=a^{2n}+\sum_1^{2n-1}(-1)^ka^{2n-k}b^k+\sum_0^{2n-2}(-1)^ka^{2n-1-k}b^{k+1}-b^{2n}$$

Now shifting indicies on the second sum, and rearranging:

$$\implies a^{2n}-b^{2n}+\sum_1^{2n-1}(-1)^ka^{2n-k}b^k+\sum_1^{2n-1}(-1)^{k-1}a^{2n-k}b^{k}$$

Then pulling out a $-1$ from the second sum we see that the two sums kill each other and we are left with $a^{2n}-b^{2n}$. It seemed like a reasonable proof to me but is lacking rigor apparently because I did not show that it holds for $n+1$. However, showing the proof for an arbitrary $n$ seemed good enough to me as we can pick any $n$ and know that there is divisibility. What am I missing? Why is showing a proof for an arbitrary $n$ not good enough? Is induction the only way to prove this?

• Looks reasonable to me. – rogerl Jun 10 '14 at 0:42
• Why do you think so? Because of the reason I stated? @rogerl – ClassicStyle Jun 10 '14 at 1:04
• For your argument to be completely rigorous, one must already have a rigorous definition of the summation operator. It can be defined by recursion. – Bill Dubuque Jun 10 '14 at 3:22

The polynomial reminder theorem states that $(x-c)$ is a divisor of a polynomial $f(x)$ if and only if $f(a)=0$. Let $x=a$, $f(a)=a^{2n}-b^{2n}$ and $c=-b$. Therefore $(a+b)$ is a divisor of $f(a)$ if and only if $f(-b)=0$. $f(-b)=(-b)^{2n}-b^{2n}=0$. So therefore $(a+b)$ divides $a^{2n}-b^{2n}$.