0

proof:

Suppose that $3| x^3+2x+1$ and $x$ is a rational number, $\frac{p}{q}$, $gcd(p, q) = 1, q \ne 0$

sub $\frac{p}{q}$ into $x^3+2x+1$:

$\frac{p^3+2pq^2+q^3}{q^3} =3d$ for some integer d

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

this means that,

$3|(p^3+2pq^2+q^3)$

I got stuck from this point onwards and could not find a contradiction, any hints on how should I proceed with the proof?

Sam Jing Wen
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3 Answers3

2

It should be $3\mid (p^3+2pq^2+q^3)$.

If $p\equiv 0\pmod 3 $ then we must have $3\mid q^3$. As $3$ is prime this implies that $3\mid q$, contradicting the fact that $\gcd(p,q)=1$.

If $p\equiv 1\pmod 3$ we must have $1+2q^2+q^3\equiv 0 \pmod 3$. Both $q\equiv 0\pmod 3$ and $q\equiv 1\pmod 3$ would lead to $1\equiv 0\pmod 3$ and $q\equiv 2\pmod 3$ would mean $2\equiv 0 \pmod 3$, so this cannot be the case either.

If $p\equiv 2\pmod 3$ necessarily $3\mid 2+q^2+q^3$. As before, all three possibilities for $q$ lead to contradictions.

Thus, $x$ may not be expressed as an irreducible fraction and therefore must be irrational.

Julio Puerta
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  • I would use the lemma $x^3 \equiv x \pmod 3.$ Then $p^3 + 2 p^2 q + q^3 \equiv 0 \pmod 3$ becomes $p + 2 p^2 q + q \equiv 0 \pmod 3$ and $p + (1 +2 p^2 ) q \equiv 0 \pmod 3.$ Finally, either $p \equiv 0$ or $ 1 + 2 p^2 \equiv 0$ – Will Jagy Mar 30 '24 at 16:48
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    I agree that this is faster, but due to the small number of cases it doesn't affect that much. – Julio Puerta Mar 30 '24 at 16:53
1

$x$ cannot be an integer because we have $x^3+2x+1=3a$ where $a$ is an integer and $x^3+2x+1\equiv1\pmod3$. Suppose $x=\dfrac pq$ with $(p,q)=1$ so we have $$\frac{p^3+2pq^2}{q^3}=(3a-1)\in\Bbb Z$$ This implies $$p^3\equiv0\pmod q\iff p^3=mq \text { where m is an integer }$$ But then all prime factor of $q$ divides $p^3$ which contradicts that $p$ and $q$ are coprimes. Then $x$ should be irrational.

Piquito
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0

Assume $q$ is rational such that $q^3 + 2q + 1=3n$, then $x^3+2x+1-3n$ has a rational root $q$. By rational root theorem only possible denominators of $q$ are $\pm 1$, hence $q$ is an integer. But as noted $q^3+2q+1-3n \equiv 1 \pmod 3$, in particular $q^3+2q+1-3n\neq 0$, a contradiction.

Sil
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