factorization-of-an-plus-minus-zn

Factorization of $a^n\pm z^n$

Summary: §150–§153: closed-form decomposition of $a^n+z^n$, $a^n-z^n$, and $a^{2n}-2a^n{z}^n\cos g+z^{2n}$ into real linear and trinomial factors, with the cosines indexed by equally spaced arcs.

Sources: chapter9

Last updated: 2026-04-29

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$a^n+z^n$ (§150)

Apply the §148 method (see trinomial-factor) with $\alpha =a^n$, only $\zeta z^n$ nonzero (coefficient $1$). Setting $r=p/q$, the two real equations become

$0=a^n+r^n\cos n\varphi ,0=r^n\sin n\varphi .$

The second forces $n\varphi =k\pi$. Among integer $k$, the value of $\cos n\varphi$ is either $+1$ (if $k$ is even, $n\varphi =2k\pi$) or $-1$ (if $k$ is odd, $n\varphi =(2k+1)\pi$). The first equation $a^n+r^n\cos n\varphi =0$ requires $\cos n\varphi =-1$, hence

$n\varphi =(2k+1)\pi ,\varphi =\frac{(2k+1)\pi }{n},r=a.$

So $p=a$, $q=1$, and the trinomial factor of $a^n+z^n$ is

$\boxed{a^2-2az\cos \frac{(2k+1)\pi }{n}+z^2}$

Substituting $2k+1=1,3,5,\dots$ produces every factor; values past $n$ repeat because $\cos (2\pi \pm \varphi )=\cos \varphi$ (source: chapter9, §150).

If $n$ is odd, choosing $2k+1=n$ gives $\varphi =\pi$ and the factor $a^2+2az+z^2=(a+z{)}^2$. Take only the square root: $a+z$ is the real linear factor. So

$n$	$a^n+z^n$ factors
$1$	$a+z$
$2$	$a^2+z^2$
$3$	$(a+z)(a^2-2az\cos (\pi /3)+z^2)$
$4$	$(a^2-2az\cos (\pi /4)+z^2)(a^2-2az\cos (3\pi /4)+z^2)$
$5$	$(a+z)(a^2-2az\cos (\pi /5)+z^2)(a^2-2az\cos (3\pi /5)+z^2)$
$6$	$(a^2-2az\cos (\pi /6)+z^2)(a^2-2az\cos (3\pi /6)+z^2)(a^2-2az\cos (5\pi /6)+z^2)$

(source: chapter9, §150 examples).

$a^n-z^n$ (§151)

The same calculation, but now the first equation $a^n-r^n\cos n\varphi =0$ requires $\cos n\varphi =+1$, hence $n\varphi =2k\pi$, $\varphi =2k\pi /n$, $r=a$. The trinomial factor is

$\boxed{a^2-2az\cos \frac{2k\pi }{n}+z^2}$

with $2k=0,2,4,\dots$ up to $n$. At $2k=0$ the factor degenerates to $a^2-2az+z^2=(a-z{)}^2$, so $a-z$ is a real linear factor. If $n$ is even, $2k=n$ gives $a^2+2az+z^2=(a+z{)}^2$, so $a+z$ is also a real linear factor (source: chapter9, §151).

$n$	$a^n-z^n$ factors
$1$	$a-z$
$2$	$(a-z)(a+z)$
$3$	$(a-z)(a^2-2az\cos (2\pi /3)+z^2)$
$4$	$(a-z)(a+z)(a^2-2az\cos (2\pi /4)+z^2)$
$5$	$(a-z)(a^2-2az\cos (2\pi /5)+z^2)(a^2-2az\cos (4\pi /5)+z^2)$
$6$	$(a-z)(a+z)(a^2-2az\cos (2\pi /6)+z^2)(a^2-2az\cos (4\pi /6)+z^2)$

These tables are the cyclotomic factorization, written in terms of cosines rather than primitive roots of unity. Modern notation: the roots of $z^n-a^n=0$ are $z=a{\zeta }^k$ with $\zeta =e^{2\pi i/n}$, and $z=a\cos (2k\pi /n)+ia\sin (2k\pi /n)$.

$a^{2n}-2a^n{z}^n\cos g+z^{2n}$ (§152–§153)

This expression — the product of two complex factors $a^n-z^n{e}^{ig}$ and $a^n-z^n{e}^{-ig}$ — has no factor of the simple form $\eta +\theta z^n$. Apply §149 with $m=2n$ to obtain $r=a$ and $\sin 2n\varphi =2\cos g\sin n\varphi$, hence $\cos n\varphi =\cos g$ and $n\varphi =2k\pi \pm g$. The general trinomial factor is

$\boxed{a^2-2az\cos \frac{2k\pi \pm g}{n}+z^2}$

over $2k=0,2,4,\dots \le n$ (source: chapter9, §153).

Examples:

• $n=1$: $a^2-2az\cos g+z^2$ has the single factor $a^2-2az\cos g+z^2$.
• $n=2$: $a^4-2a^2{z}^2\cos g+z^4=(a^2-2az\cos (g/2)+z^2)(a^2+2az\cos (g/2)+z^2)$.
• $n=3$: $a^6-2a^3{z}^3\cos g+z^6$ splits into three trinomials with arcs $g/3$, $(2\pi -g)/3$, $(2\pi +g)/3$.

The case $g=0$ recovers $(a^n-z^n{)}^2$ — square root gives $a^n-z^n$ — and the case $g=\pi$ recovers $(a^n+z^n{)}^2$ giving $a^n+z^n$. Setting $g=2\pi k/n$ recovers the §150–§151 results, so this is the master formula.

Why these matter for what follows

§154 observes that any polynomial in $z^n$ — for instance $\alpha +\beta z^n+\gamma z^{2n}$ — first factors over $\mathbb{R}$ into pieces of the forms just handled, and each piece then splits into trinomials by the formulas above. So every polynomial admits a constructive factorization into real linear and trinomial factors, fulfilling the §32 promise (see fundamental-theorem-of-algebra).

§155 onwards extends the same trick to infinite series. Treating $e^x$, $\sin x$, $\cos x$ as polynomials of “infinite degree”, Euler applies the §150–§153 formulas with $n$ infinite to obtain the infinite-product expansions of sine-infinite-product, cosine-infinite-product, and the exponential family.

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• trinomial-factor
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• fundamental-theorem-of-algebra
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• chapter-9-on-trinomial-factors