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Every real solution of $x^{3}+x+3=0$ is irrational.

I saw a similar post, but none of the comments followed my approach, so I was wondering if it is valid.

My attempt:

For the sake of contradiction, suppose it is not the case that every real solution of $x^{3}+x+3=0$ is irrational. Thus, there is a solution of $x^{3}+x+3=0$ that is rational. Let this solution be $u/v$, where $u,v\in\mathbb{Z}$, $v\neq0$, and $u/v$ is fully reduced, that is, $\textrm{gcd}(u,v)=1$.

Since $u/v$ is a solution to $x^{3}+x+3=0$, it follows that $(u/v)^{3}+(u/v)+3=0$. Thus, \begin{align} \left(\frac{u}{v}\right)^{3}+\left(\frac{u}{v}\right)+3&=0\nonumber\\ \left(\frac{u^{3}}{v^{3}}\right)+\left(\frac{u}{v}\right)&=-3\nonumber\\ \frac{u^{3}+uv^{2}}{v^{3}}&=-3\nonumber\\ u^{3}+uv^{2}&=-3v^{3}\nonumber\\ u(u^{2}+v^{2})&=-3v^{3}\nonumber \end{align} Now, we look into the equation above modulo $2$. Since every integer is congruent to $0$ or $1$ modulo $2$, we produce the following table to account for all possibilities:

\begin{array}{c|c|c|c|c||c|c} u(\textrm{mod} 2) & v(\textrm{mod} 2) & u^{2}(\textrm{mod} 2) & v^{2}(\textrm{mod} 2) & v^{3}(\textrm{mod} 2) & u(u^{2}+v^{2})(\textrm{mod} 2) & -3v^{3}(\textrm{mod} 2)\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 & 1 & 2\equiv 0 & 1\\ \end{array}

As we can see, the left-hand side is congruent to the right-hand side modulo $2$ only when $u\equiv0 \pmod 2$ and $v\equiv0 \pmod 2$. Hence, $u$ and $v$ are even, which contradicts the fact that $u/v$ is fully reduced. $\blacksquare$

My questions are,

  • Can I claim that if $a\not\equiv b(\textrm{mod } n)$ then $a\neq b$ without proving it?
  • Also, is this table method ok to use?

I have never seen it done like this before, so I don’t know if it is correct or standard. (I am self-teaching math, by the way)

Martin Sleziak
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Igor Mello
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    A shorter proof consists of using the rational root test. – Dietrich Burde Jul 01 '22 at 13:28
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    About your question in the last paragraph, the contrapositive would be $a=b \implies a\equiv b \pmod n$. It's clear that as $a=b$, they have the same remainder when divided by $n$, or $n\mid a-b$. – peterwhy Jul 01 '22 at 13:35
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    Your proof is fine, if a little long. With $u^3+uv^2+3v^3$ it is easy to see that if $u$,$v$ are both odd then all three terms are odd, and if one of $u$,$v$ is odd and the other even then exactly one of the terms is odd. Either way, the sum of the three terms is odd so cannot be zero. – Jaap Scherphuis Jul 01 '22 at 13:42
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    +1: (also) to your posting. Very nice work done. Independent of the quality of your posting, I (also) agree that your proof is valid. I also think that the general methods that you conjured in your posting may be similarly applied in a wide range of Number Theory problems. In fact, my understanding is that in many such problems, the methods used in your posting represent the standard line of attack. – user2661923 Jul 01 '22 at 14:17
  • Special case of the Parity Root Test in the linked dupes. – Bill Dubuque Jul 01 '22 at 14:36
  • Optimization: as explained here we may assume wlog that $u$ and $v$ have opposite parity (by cancelling $2$'s from $u/v$, i.e. $u/v$ is $2$-reduced, i.e. in $2$-lowest terms) $\ \ $ – Bill Dubuque Jul 01 '22 at 14:39
  • The key idea of the argument is that integer roots of integer coef polynomials persist $!\bmod n$ (cf. modular root test) and this does require justification (at this level). – Bill Dubuque Jul 01 '22 at 14:54
  • Note that the proof also works for complex Gaussian integers $,a+b,i,\ a,b\in \Bbb Z,$ since they too have a sense of parity as explained in the first linked dupe. – Bill Dubuque Jul 01 '22 at 15:04
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    @BillDubuque, How can a solution-verification question be closed as a dupe? OP doesn't want to know how to prove it; they want to know if their own proof is valid. – Peter Phipps Jul 01 '22 at 18:15
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    @Peter The OP's question(s) are answered by combining the above comments and linked answers. There is nothing novel here. – Bill Dubuque Jul 02 '22 at 11:29
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    @PeterPhipps: and I have no idea how this question is a duplicate to "Do odd imaginary numbers exist?" –  Jul 02 '22 at 15:34
  • @user1046533 Nobody claimed that the questions are the same. Rather, the above comment asserts it is a duplicate of the linked answers on the Parity Root Test (cf. the SE closure banner: "This question already has answers here"). It is often the case that common theorems/answers serve to answer many variants of questions. – Bill Dubuque Jul 02 '22 at 15:47
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    @Bill I think such a close destroys the purpose of the site. This is not mathOverflow. "The OP's question(s) are answered by combining the above comments and linked answers" But comments are not supposed to be answers. –  Jul 02 '22 at 15:48
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    And your so-called "variant" is way too broad. Most elementary questions on this site are all, in some very broad sense, nothing but a few tricks in undergraduate courses. It is because people who address the nuances make this Q&A site useful. If one is going to make it a MSE-pedia site, I am afraid the essential function of this site is undermined. –  Jul 02 '22 at 15:53
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    @user1046533 There is no value added to the site by an "answer" saying "Yes, your proof is corrrect" when there are already (many) correct proofs that have been evaluated by the community. Further, leaving it open almost always results in users posting many duplicate answers, which leads to rampant duplication - making it difficult if not impossible to locate "best" answers. This is why many users believe SV questions should be banished. – Bill Dubuque Jul 02 '22 at 15:53

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