Sum of Ordinate Powers Curves

Summary: §§367–373 of chapter-16-on-finding-curves-from-properties-of-the-ordinate. Given the symmetric condition $P{M}^n+P{N}^n=a^n$ on the two ordinates of a two-valued curve $y^2-Py+Q=0$, Newton’s identities pin down one polynomial relation in $P,Q$. The cleanest closed form is the master equation $(P-y{)}^n+y^n=a^n$, expressing the fact that if $y$ is one ordinate then $P-y$ is the other. An alternate form $y^n+(Q/y{)}^n=a^n$ uses the product instead. The equations extend to negative-integer and fractional exponents.

Sources: chapter16 (§§367–373). Figure 78 (in figures76-80) for the setup.

Last updated: 2026-05-11.

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The problem (§367)

For each abscissa $AP=x$, let $PM,PN$ be the two ordinates (figure 78). We want curves with

$P{M}^n+P{N}^n=a^n$

for $a$ constant. Since $PM+PN=P$ and $PM\cdot PN=Q$, Newton’s identities convert the condition to a polynomial relation in $P,Q$:

$n$	Newton identity for $P{M}^n+P{N}^n$	Required condition
1	$P$	$P=a$ (the diameter case, two-ordinate-sum-and-product)
2	$P^2-2Q$	$P^2-2Q=a^2$, i.e., $Q=(P^2-a^2)/2$
3	$P^3-3PQ$	$P^3-3PQ=a^3$, i.e., $Q=(P^3-a^3)/(3P)$
4	$P^4-4P^{2}Q+2Q^2$	$P^4-4P^{2}Q+2Q^2=a^4$
$n$	(Newton expansion)	one polynomial relation linking $P,Q$

Solving for $Q$ in terms of $P$ gives the curve equation $y^2-Py+Q(P,a)=0$, with $P$ still a free function of $x$.

The $n=2$ ellipse (§367)

With $Q=(P^2-a^2)/2$, choose $P=2nx$ (linear in $x$). The curve is

$y^2-2nxy+2n^2{x}^2-\frac{1}{2}{a}^2=0,$

a conic. Solving the quadratic gives $y=nx\pm \sqrt{a^2/2-n^2{x}^2}$. The radicand is bounded, so the curve is an ellipse, centered at the origin (since the $x\mapsto -x$ symmetry exchanges branches via reflection). This is exactly the ellipse $P{M}^2+P{N}^2=a^2$ — its two ordinates have constant sum of squares, an instance of ellipse-sum-of-squares-conjugate-diameter.

The $n=3$ cubic (§369)

With $Q=(P^3-a^3)/(3P)$, the curve equation is

$y^2-Py+\frac{P^3-a^3}{3P}=0,$

or after clearing the denominator and choosing $P=3nx$, $a=3nb$:

$x{y}^2-3n{x}^{2}y+3n^2{x}^3-3n^2{b}^3=0.$

This is a third-order line. Euler notes it belongs to the second species of the cubic classification (see cubic-species-classification).

The $n=4$ irrationality (§370)

For $n=4$, the relation $P^4-4P^{2}Q+2Q^2=a^4$ solved for $Q$ gives

$Q=P^2\pm \sqrt{P^4/2+a^4/2}.$

The radical $\sqrt{P^4/2+a^4/2}$ is generally irrational, but $Q$ was required to be a non-irrational function of $x$. Euler’s resolution: from $y=P/2\pm (-3P^2/4\pm \sqrt{P^4/2+a^4/2}{)}^{1/2}$, the value of $y$ cannot be real unless the radical is taken with the positive sign. With this sign choice, $y$ has only two values (despite $Q$ being double-valued formally), as the problem requires.

The general master equation (§371)

The closed-form pattern for all $n$ is

$\boxed{(P-y{)}^n+y^n=a^n}$

with $P$ any single-valued function of $x$ chosen freely. The argument is immediate: if $y$ is one of the two ordinates, then $P-y$ is the other (since they sum to $P$). The condition $P{M}^n+P{N}^n=a^n$ thus reads $y^n+(P-y{)}^n=a^n$ directly.

For $n=5$: $P^5-5P^{3}Q+5P{Q}^2=a^5$; since $Q=-y^2+Py$, the equation expands to $P^5-5P^{4}y+10P^3{y}^2-10P^2{y}^3+5P{y}^4=a^5$, which is $(P-y{)}^5+y^5=a^5$. Similarly $(P-y{)}^6+y^6=a^6$ for $n=6$, and so on.

The alternate form via $Q$ (§372)

Eliminating $P$ instead of $Q$: since $P=(y^2+Q)/y$, substitute into $P{M}^n+P{N}^n=a^n$ with $Q/y$ as the other ordinate to get

$y^n+\frac{Q^n}{y^n}=a^n,$

equivalently $y^{2n}-a^n{y}^n+Q^n=0$. Solving the quadratic in $y^n$,

$y={\left(\frac{1}{2}{a}^n\pm \sqrt{\frac{1}{4}{a}^{2n}-Q^n}\right)}^{1/n}.$

This is two-valued for each $x$ as required, provided $Q^n$ is non-irrational. The advantage of the $(P-y{)}^n+y^n=a^n$ form is lower degree.

Tables for integer and fractional exponents (§373)

The chapter tabulates the cases for negative-integer reciprocal sums:

Required property	Equation (using $P$)	Equation (using $Q$)
$1/PM+1/PN=1/a$	$aP=Py-y^2$	$aQ+a{y}^2=Qy$
$1/P{M}^2+1/P{N}^2=1/a^2$	$a^2{y}^2+a^2(P-y{)}^2=y^2(P-y{)}^2$	$a^2{Q}^2+a^2{y}^4=Q^2{y}^2$
$1/P{M}^3+1/P{N}^3=1/a^3$	$a^3{y}^3+a^3(P-y{)}^3=y^3(P-y{)}^3$	$a^3{Q}^3+a^3{y}^6=Q^3{y}^3$

And for fractional exponents:

Required property	Master equation	Rationalized form
$\sqrt{PM}+\sqrt{PN}=\sqrt{a}$	$\sqrt{y}+\sqrt{P-y}=\sqrt{a}$, or $y=\sqrt{ay}-\sqrt{Q}$	$y^2-Py+\frac{1}{4}(a-P{)}^2=0$, or $y^2-(a-2\sqrt{Q})y+Q=0$
$P{M}^{1/3}+P{N}^{1/3}=a^{1/3}$	$y^{1/3}+(P-y{)}^{1/3}=a^{1/3}$	$y^2-Py+\frac{1}{27a}(a-P{)}^3=0$, or $y^2-(a-3(aQ{)}^{1/3})y+Q=0$

The general claim is: one equation $(P-y{)}^n+y^n=a^n$ covers all cases, whether $n$ is a positive integer, negative integer, or fraction. Equivalent forms in $y,Q$ also exist.

Why the master equation is so clean

The clean form $(P-y{)}^n+y^n=a^n$ packages three ingredients:

1. Vieta — the two ordinates are $y$ and $P-y$, dual roles played by the two terms.
2. No symmetric function machinery needed — the condition reads directly without Newton’s identities.
3. Universality in $n$ — the same equation works for integer, fractional, and negative exponents.

This is the chapter’s prototype of “right-side-of-the-equation thinking”: rather than expanding to $P,Q$ via identities, encode the condition $f(y_1,y_2)=$ const by substituting $y,P-y$ for the two roots.

Figures

Figures 76–80

Related pages

• chapter-16-on-finding-curves-from-properties-of-the-ordinate — chapter summary.
• two-ordinate-sum-and-product — the $n=1$ (sum) and product cases.
• ellipse-sum-of-squares-conjugate-diameter — geometric meaning of the $n=2$ case for the ellipse.
• cubic-species-classification — the $n=3$ curve lives in the second species.
• chord-rectangle-property — chapter 5 forward-direction prototype.