ch2.0.3-compound-indeterminate-equations

Ch2.0.3 — Of Compound Indeterminate Equations in which One Unknown does not exceed the First Degree

Summary: Treats indeterminate equations containing products $xy$ or higher powers of $x$, where $y$ appears only to the first degree; the method reduces each case to a divisibility condition on a known constant.

Sources: chapter-2.0.3

Last updated: 2026-05-04

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General Form

Equations of this class have the general shape

$a+bx+cy+d{x}^2+exy+f{x}^3+g{x}^{2}y+h{x}^4+k{x}^{3}y+\cdots =0$

Because $y$ appears only linearly, it can be isolated; but the resulting expression for $y$ contains $x$ in the denominator, so integrality forces $x$ to be a divisor of a certain known number. (source: chapter-2.0.3, §31)

Simplest Case: $xy+x+y=c$

For $xy+x+y=c$, solving for $y$ gives

$y=\frac{c-x}{x+1}=-1+\frac{c+1}{x+1}$

so $x+1$ must divide $c+1$. Each divisor of $c+1$ yields a valid $x$, and then $y$ follows. (source: chapter-2.0.3, §32)

Example (§32): $xy+x+y=79\Rightarrow y=-1+\frac{80}{x+1}$. The divisors of 80 give $x=0,1,3,4,7,9,15,19,39,79$; symmetry halves this to five genuine pairs.

General Linear-Product Case: $xy+ax+by=c$

Solving for $y$:

$y=\frac{c-ax}{x+b}=-a+\frac{ab+c}{x+b}$

so $x+b$ must divide $ab+c$. Each divisor $d$ of $ab+c$ gives $x=d-b$ and $y=\frac{ab+c}{d}-a$; the roles of the two factors are interchangeable, so each factorisation $ab+c=fg$ yields two solution pairs: $(x,y)=(f-b,g-a)$ and $(g-b,f-a)$. (source: chapter-2.0.3, §33)

Most General First-Degree Case: $mxy=ax+by+c$

Solving for $y$:

$my=\frac{max+mc}{mx-b}=a+\frac{mc+ab}{mx-b}$

so $mx-b$ must divide $mc+ab$. Only those divisors of $mc+ab$ that are congruent to $-b(\mathrm{mod}m)$ (equivalently, divisors $d$ such that $d+b\equiv 0(\mathrm{mod}m)$) yield integer $x=\frac{d+b}{m}$. (source: chapter-2.0.3, §34)

Example (§34): $5xy=2x+3y+18\Rightarrow 5y=2+\frac{96}{5x-3}$. Divisors of 96 that when added to 3 give a multiple of 5 are $\{2,12,32\}$, yielding $(x,y)\in \{(1,10),(3,2),(7,1)\}$.

Structural Observation on Divisors

In the general case $my-a=\frac{mc+ab}{mx-b}$, if a number of the form $mc+ab$ has a divisor of the form $mx-b$, the complementary factor is automatically of the form $my-a$. Thus $mc+ab$ can be written as $(mx-b)(my-a)$, a factorisation into two linear forms. This duality is of fundamental importance in number theory. (source: chapter-2.0.3, §35)

Quadratic Terms: $xy+x^2=2x+3y+29$

When $x^2$ appears alongside $xy$, isolating $y$ and performing polynomial division still yields a divisor condition. (source: chapter-2.0.3, §36)

Example (§36): $xy+x^2=2x+3y+29\Rightarrow y=-x-1+\frac{26}{x-3}$, so $x-3$ must divide 26. Divisors $1,2,13,26$ give $x=4,5,16,29$, but $x=16$ forces $y=-15$ (rejected) and $x=29$ is similarly excluded.

Preview: Rationalization of Surds

When $y$ is also of degree $\ge 2$, solving for $y$ introduces radical expressions in $x$. The central problem then becomes choosing $x$ so that those radicals become rational integers. Euler states that “the great art of Indeterminate Analysis consists in rendering those surd, or incommensurable formulas rational” — the subject of the chapters that follow. (source: chapter-2.0.3, §37)

Practice Questions (Lagrange additions)

Ten exercises are appended, including:

• $24x=13y+16$ — integer solution $x=5,y=8$
• $87x+256y=15410$ — least $x=30$, greatest $y=12800$
• Number of integer solutions to $5x+7y+11z=224$ — answer 60
• Paying £100 in guineas (21s 6d) and pistoles (17s) — three ways
• LCM of $1,2,\dots ,9$ — answer 2520

Related pages

• indeterminate-analysis
• linear-diophantine-equations
• greatest-common-divisor
• ch2.0.1-indeterminate-equations-first-degree
• ch2.0.2-regula-caeci