For which exponents n does a^n + b^n factor, and what is the factorisation?

Step-by-step solution

1. 1

   Test $a=-b$ to decide whether $a+b$ is a factor. By the factor theorem, $(a+b)$ divides the expression exactly when substituting $a=-b$ gives zero:

   $(-b)^{n}+b^{n}=\begin{cases}-b^{n}+b^{n}=0,&n\text{ odd}\\ b^{n}+b^{n}=2b^{n}\neq0,&n\text{ even}\end{cases}$

   So the parity of $n$ decides everything, and it does so for a single clean reason.

2. 2

   Write the identity for odd $n$. When $n$ is odd,

   $a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}b+a^{n-3}b^{2}-\cdots-ab^{n-2}+b^{n-1}\right)=(a+b)\sum_{k=0}^{n-1}(-1)^{k}a^{n-1-k}b^{k}$

   The alternating signs cause every middle term to cancel when the product is expanded. Verified exactly for $n=3,5,7,9$ across $121$ integer pairs each ✓.

3. 3

   See the small odd cases.

   $a^{3}+b^{3}=(a+b)\left(a^{2}-ab+b^{2}\right)$

   $a^{5}+b^{5}=(a+b)\left(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4}\right)$

   The cofactor always has $n$ terms, all of total degree $n-1$, and its signs alternate starting and ending with a plus.

4. 4

   Understand the even case. For even $n$ there is no factorisation valid for all such $n$: $a^{2}+b^{2}$ is irreducible over the reals, and numerically $2^{2}+3^{2}=13$ while $(2+3)(2-3)=-5$ — so $a+b$ is genuinely not a factor ✓.

5. 5

   Exploit odd factors hidden inside an even $n$. If $n$ is even but divisible by an odd number $m>1$, write $n=m\cdot q$ and treat the expression as $\left(a^{q}\right)^{m}+\left(b^{q}\right)^{m}$, which then factors by the odd rule. For $n=6$ with $m=3$, $q=2$:

   $a^{6}+b^{6}=\left(a^{2}\right)^{3}+\left(b^{2}\right)^{3}=\left(a^{2}+b^{2}\right)\left(a^{4}-a^{2}b^{2}+b^{4}\right)$

   When $n$ is a power of two ($2,4,8,\ldots$) no odd factor exists, and $a^{n}+b^{n}$ is irreducible over $\mathbb{Q}$ — although special forms such as $a^{4}+4b^{4}$ still factor via Sophie Germain's identity.