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Considering the assertion: The polynomial $x^3 - 3x + 1$ has no rational roots, the following is a proof by contradiction:

Let a root of $x^3 - 3x + 1$ be written in the form of $\frac{p}{q}$ where $p$ and $q$ are coprime.

Thus $\frac{p^3}{q^3}$ - $\frac{3p}{q} + 1 = 0$

$\frac{p^3}{q^3}$ - $\frac{3p}{q} = -1$

By multiplying both sides by $q^3$ we have

$p^3 -3q^2p = -q^3$

$p^3 + q^3 = 3q^2p$

The above equation only holds true if both p and q are even, which I have obtained by trying all 4 cases.

Case 1

p = odd, q = even.

$odd + even \neq 3*odd*even^2 \neq even$

Case 1 is false.

Case 2

p = even, q = odd.

$even + odd \neq 3*even*odd^2 \neq even$

Case 2 is false.

Case 3

p = odd, q = odd.

$odd + odd \neq 3*odd*odd^2 \neq odd$

Case 3 is false.

Case 4

p = even, q = even.

$even + even = 3*even*even^2 = even$

Case 4 is true.

Now we arrive at the contradiction of the initial condition that $p$ and $q$ are coprime. Thus the polynomial $x^3 - 3x + 1$ has no rational roots.

The question is, what is the proper way to present such a proof? It seems like writing out all the cases and treating the words odd and even as pseudo variables is a very crude and improper and unofficial way of presenting a proof.

I'm looking for some kind of algebraic presentation that proves $p$ and $q = 2k$ for some $k \in \mathbb Z $

Update

Ok maybe I wasn't very clear in stating what I'm asking for. I'm looking for an algebraic way of proving that $p^3 + q^3 = 3q^2p$ ONLY holds true when both $p$ and $q$ are even numbers, very much unlike the makeshift proof I did above by listing out all 4 cases and checking them one by one.

Joffan
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Lagrangian
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    To prove that your set of cases is exhaustive, and to analyze each case is called "brute force." Your approach is rigorous even if it is perhaps not as elegant as you would like. – Doug M Jun 29 '16 at 16:13
  • Case 4 doesn't really lead to a contradiction. You just should say: this case can't happen since we supposed $p$ and $q$ are coprime. That said, a shorter version proves $p$ is a divisor of the constant term, and $q$ a divisor of the leading coefficient, using Gauß'lemma, hence the possibiliites are only $\pm1$, none of which is a root. – Bernard Jun 29 '16 at 16:14

4 Answers4

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You can use Rational roots theorem to check that if ${p\over q}$ is a rational root in its lowest form with $p\in\mathbb Z,\;q\in\mathbb N\,$, $$ p|1,\;q|1\implies p=\pm1,q=1\implies{p\over q}=\pm1 $$ Plugging in $x=\pm1$ we see $y(1)=-1,y(-1)=3$. Therefore no rational root exists.

Jack's wasted life
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from here: $p^3 -3q^2p = -q^3$

I would say that the expression on the left is divisible by $p.$ Which means that the expression on the right is divisible by $p.$ Since $p$ and $q$ are co-prime $p = 1.$

Doug M
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To undertake your even/odd analysis with a different layout, I would say:

Working $\bmod 2$,

$$p^3 + q^3 \equiv \begin{cases} 0 & p\equiv q \\ 1 & p \not\equiv q \end{cases}$$

and

$$3q^2p \equiv \begin{cases} 1 & p\equiv q \equiv 1\\ 0 & \text{otherwise} \end{cases}$$

Thus only $p\equiv q\equiv0 \bmod 2$ allows $p^3 + q^3 = 3q^2p$, which contradicts the construction that $p$ and $q$ are in lowest terms, as both are divisible by $2$.

Joffan
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Example of a different style:

"Case 1. If $p$ is odd and $q$ is even then $p^3$ is odd and $q^3$ is even, so $p^3+q^3$ is odd but $3pq^2$ is even, so $p^3+q^3\ne 3pq^2.$"

Your proof is valid. There is nothing illogical about a case-by-case method.

However we could say $$p^3+q^3=3pq^2\implies p^3+q^3\equiv 3pq^2 \pmod 2\implies$$ $$\implies p+q\equiv pq \pmod 2\implies 1\equiv (p-1)(q-1)\pmod 2\iff$$ $$\implies p-1\not \equiv 0 \not \equiv q-1\pmod 2 \implies$$ $$\implies p\equiv q\equiv 0\pmod 2.$$

....and therefore $\neg (p\equiv q\equiv 0\pmod 2)\implies p^3+q^3\ne 3pq^2.$

DanielWainfleet
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