3
$\begingroup$

There is a problem:

Is there any triple of rational numbers $(a,b,c)$ such that $a+b+c=0$ and $abc=1$?

My answer is $\boxed{No}$. Here is my proof:

Suppose such a triple exists. Then it must consist of the $3$ solutions (counting multiplicities) of the cubic equation: $$X^3+kX-1=0,$$ where $k$ is a rational number and $k\ne 0$ (because $k=0$ makes the equation get non-rational solutions). This equation has two roots: $$X_{1,2}=\frac{1}{2}\pm \sqrt{\frac{1}{4}+\frac{k^3}{27}}.$$ Thus $\frac{1}{4}+\frac{k^3}{27}=m^2$ with some rational number $m$. By a linear change of variables, the result is equivalent to the Elliptic curve $$(E): y^2=x^3+\frac{1}{4}$$ having a non-trivial rational point. But $(E)$ has rank $0$ and torsion group $\mathbb{Z}/3\mathbb{Z}$ (checked by CoCalc) so $E(\mathbb{Q})=\{O, (0,\frac{1}{2}),(0,-\frac{1}{2})\}$, a contradiction. This proved the claim.

Did I make a mistake? I am still looking for an elementary solution.

$\endgroup$
6
  • 1
    $\begingroup$ How did you get those two(?) solutions to the cubic? They don't seem to be correct. For example, for $k>0$ the polynomial function is clearly increasing, so there is only one real solution. $\endgroup$ Commented Sep 4, 2025 at 15:22
  • $\begingroup$ For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. $\endgroup$ Commented Sep 4, 2025 at 15:30
  • $\begingroup$ Did you use Cartan’s algorithm for cubic polynomials?. I would have thought you always have an odd number of roots for a cubic $\endgroup$ Commented Sep 4, 2025 at 15:49
  • $\begingroup$ Cardan* not Cartan $\endgroup$ Commented Sep 4, 2025 at 15:54
  • 1
    $\begingroup$ I show below that your problem is logically equivalent to the exponent-3 case of Fermat's Last Theorem. $\endgroup$ Commented Sep 6, 2025 at 0:46

3 Answers 3

Reset to default
15
$\begingroup$

I have for you what I feel is a beautiful proof. Necessarily two of them must be negative, say $b$ and $c$ are negative and $a=-(b+c)\gt0$.

It follows that $$abc=1\iff bc(b+c)=-1$$ I have used some time ago (doing my thesis) this identity, which you can verify: $$\boxed{[9bc^2+(c-b)^3]^3+[9b^2c-(c-b)^3]^3=bc(b+c)[3(b^2+bc+c^2)]^3}$$ Thus, if $bc(b+c)=-1$, because $-1$ is a cube, we would have it is impossible by Fermat's last theorem for the exponent $3$.

$\endgroup$
9
  • 1
    $\begingroup$ You cannot have one negative and two positive as $abc$ would then be negative and so not $1$. $\endgroup$ Commented Sep 4, 2025 at 17:07
  • 2
    $\begingroup$ But I do like the use of Fermat's last theorem for $n=3$ $\endgroup$ Commented Sep 4, 2025 at 17:13
  • 2
    $\begingroup$ @Tung N. Dinh. That identity show that all integer of the form $ab(a+b)$ is representable as a sum of two cubes of rational. As an historical note, Legendre believed that $6$ is not representable as this form which is wrong because $6=2\times1(2+1)$ ; in fact the identity gives even the generator of the elliptic curve $x^3+y^3=6z^3$ which is $(37,17,21)$. $\endgroup$ Commented Sep 4, 2025 at 17:51
  • 6
    $\begingroup$ FLT only applies to non-zero terms. Is it obvious (or easy to prove) that none of those three terms is zero? $\endgroup$ Commented Sep 5, 2025 at 16:06
  • 2
    $\begingroup$ @TonyK, $3(b^2+bc+c^2)$ will never be $0$. If, for example, $9bc^2=(b-c)^3$, then dividing both sides by $c^3$ gives a cubic equation in $T=b/c$: $T^3-3T^2-6T-1=0$. By the Rational Root Theorem, this equation has no rational solutions. $\endgroup$ Commented Sep 6, 2025 at 8:16
5
$\begingroup$

Here is an elementary proof that your problem is equivalent to exponent-3 FLT.

($\Rightarrow$) Suppose such a triple $(a, b, c) \in \mathbb{Q}^3$ exists. Choose minimal positive $d \in \mathbb{N}_{>0}$ such that $da, db, dc \in \mathbb{Z}$. Then $(A, B, C) := (-da, -db, dc) \in \mathbb{Z}^3$ satisfies $C = A+B$ and $ABC = d^3$, so that $AB(A+B) = d^3 > 0$.

If a prime $p$ divides any two of $A$, $B$, and $A+B$, then it divides all three. Since $d$ is minimal, this cannot occur.

It follows that $A$, $B$, and $C = A+B$ are pairwise coprime integers multiplying to $d^3 > 0$. Thus all three are non-zero perfect cubes. $\square$

($\Leftarrow$) Suppose that $A^3 + B^3 = C^3$ with $A, B, C \in \mathbb{N}_{>0}$. Then $(a, b, c) := (-A^3/ABC, -B^3/ABC, C^3/ABC) \in \mathbb{Q}^3$ satisfies $a + b + c = 0$ and $abc = 1$. $\square$

$\endgroup$
4
$\begingroup$

Since $a+b+c = 0$ we can take $c = -(a+b)$. This turns $abc = 1$ into $a^2b+ab^2 = -1$. Solving for $a$ using the quadratic formula gives $$ a = \frac{-b^2\pm\sqrt{b^4-4b}}{2b}. $$ This will have a non-trivial solution if $b^4-4b$ is a square for some $b \neq 0$. Thus you are asking for a rational point on the curve $$ y^2 = x^4-4x $$ besides $(0,0)$. By dividing both sides by $x^4$ we get the equation $$ \left(\frac{y}{x^2}\right)^2 = -4\left(\frac{1}{x}\right)^3 + 1 $$ and by multiplying both sides by 16 we arrive at $$ \left(\frac{4y}{x^2}\right)^2 = \left(\frac{-4}{x}\right)^3 + 16. $$ By letting $Y = \left(\frac{4y}{x^2}\right)$ and $X = \left(\frac{-4}{x}\right)$ we get the elliptic curve $$ Y^2 = X^3 + 16 $$ which has rank $0$ and torsion points at $\infty,(0,\pm 4)$ which all give degenerate solutions to the original equation.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.