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In Leonhard Euler's book, "The Elements of Algebra" he seems to say that if we convert any cubic equation into the form $x^3 + ax^2 + bx + c$, and make sure that $a$, $b$ and $c$ are integer numbers, then if the equation has any rational solutions, then they can only be integer solutions. In the book:

When we have removed fractions from an equation of the third degree, according to the manner which has been explained, and none of the divisors of the last term are found to be a root of the equation, it is a certain proof, not only that the equation has no root in integer numbers, but also that a fractional root cannot exist; which may be proved as follows. Let there be given the equation $x^3 - ax^2 + b - c = 0$, in which, $a$, $b$, $c$ express integer numbers. If we suppose, for example, $x = \frac{3}{2}$, we shall have $$\frac{27}{8} - \frac{9}{4}a + \frac{3}{2}b - c = 0.$$ Now here, the first term alone has $8$ for the denominator; the others being either integer numbers, or numbers divided by $4$, or by $2$, and therefore cannot make $0$ with the first term. The same thing happens with every other fraction.

I don't really understand the argument he's trying to present. I would appreciate if someone could help me.

A rural reader
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Camelot823
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    yes, it is the rational roots theorem. If all coefficients are integers AND the highest degree coefficient is $1$ then any rational root is an integer – Will Jagy Aug 13 '23 at 00:06
  • Rational root theorem proof here: https://en.wikipedia.org/wiki/Rational_root_theorem – Ethan Bolker Aug 13 '23 at 00:21
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    without rational root theorem, just for this example, multiply the displayed equation by $4$ and move all but the first term to the right side, and get $\frac{27}{2}=9a-6b+4c,$which since $a,b,c$ are integers is saying that $\frac{27}{2}$ is an integer. – coffeemath Aug 13 '23 at 00:27

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Suppose we have a polynomial $c_nx^n + c_{n-1}x^{n-1} + \cdots + c_1x+c_0$ with integral coefficients $c_i$ and a rational root $r$. Write the rational root in reduced form, say $r = a/b$ where $a$ and $b$ are integers that are relatively prime.

Claim: $b$ is a factor of $c_n$ and $a$ is a factor of $c_0$. In particular, if $c_n = 1$ then $b = \pm 1$, so $r = a/b = \pm a$ is an integer.

Proof of claim: setting $x = r = a/b$ in the polynomial, $$ 0 = c_nx\left(\frac{a}{b}\right)^n + c_{n-1}\left(\frac{a}{b}\right)^{n-1} + \cdots + c_1\left(\frac{a}{b}\right)+c_0 $$ and now multiply through by $b^n$ to clear the denominator: $$ 0 = c_na^n + c_{n-1}a^{n-1}b + \cdots + c_1ab^{n-1} +c_0b^n. $$ This equation has no denominators anywhere: each term is an integer.

Every term except maybe $c_na^n$ has a factor of $b$, and the left side is $0$, so the term $c_na^n$ has a factor of $b$. Since $a$ and $b$ are relatively prime, the only way $b$ can be a factor of $c_na^n$ is to have $b$ be a factor of $c_n$.

Every term except maybe $c_0b^n$ has a factor of $a$, and the left side is $0$, so the term $c_0b^n$ has a factor of $a$. Since $a$ and $b$ are relatively prime, the only way $a$ can be a factor of $c_0b^n$ is to have $a$ be a factor of $c_0$.

KCd
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