I gave this problem to my undergraduate assistant, as I saw that Euler had originally solved it (although I am having trouble finding his proof). After working on it for two weeks, we boiled the hard cases down to showing that (1) in $\mathbb{Z}[w]$ the fundamental unit is $1+w+w^2$ (where $w$ is the real cube-root of 2) [which I'm sorry to say, I'm not certain I know off the top of my head how to prove], and (2) using that fundamental unit, I found a crazy ad hoc computation to show that there are only the obvious solutions.

So I'm wondering if someone else out there is more clever, knows where I can find Euler's proof, or if there is another nice elementary proof in the literature.

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    $\begingroup$ Elementary includes avoiding elliptic curves? Because there are not-so-hard arguments with the basic theory set up. $\endgroup$ Sep 22, 2010 at 0:07
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    $\begingroup$ Building on Cam McLeman's comment, this elliptic curve has a point of order 6, and hence possesses a 2-isogeny. A descent via 2-isogeny shouldn't be too hard to write out explicitly. It would be interesting (but perhaps not what Pace Nielsen is going for) if someone could give an elementary classification of the rational points on this curve based on the fact that it is the modular curve $X_0(36)$. $\endgroup$ Sep 22, 2010 at 0:25
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    $\begingroup$ You can find a discussion of two approaches to this problem in Schoof's "Catalan's Conjecture." An appendix to the book describes Euler's original approach (the reference for which Schoof includes), and Chapter 4 of the book describes an alternative approach due to William McCallum. $\endgroup$
    – user2490
    Sep 22, 2010 at 1:16
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    $\begingroup$ I sketched Euler's proof in "A note on Pépin's counter examples to the Hasse principle for curves of genus 1", which you can find online. $\endgroup$ Sep 22, 2010 at 6:00
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    $\begingroup$ Schoof's book reprints Euler's argument in the original Latin :-) $\endgroup$ Sep 22, 2010 at 16:55

4 Answers 4


If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let $u$ and $v$ be the complex roots of $z^2+z+1=0$.

Theorem. Let $A$, $B$, $C$ be non-zero elements of $\mathbb{Q}[u]$ with sum $0$ and product twice a cube. Then some two of $A$, $B$, $C$ are equal.

Corollary. Suppose $x$ is in $\mathbb{Q}$ and $x^2-1$ is a cube. Then $x$ is $1$, $-1$, $0$, $3$, or $-3$.

(To prove the corollary let $A=1+x$, $B=1-x$ and $C=-2$).

The proof of the theorem is a reductio ad absurdum. If there's a counterexample, there's one with $A$, $B$, $C$ in $\mathbb{Z}[u]$; take such a counterexample with $d=\min(|A|,|B|,|C|)$ as small as possible. Then $A$, $B$, $C$ are pairwise coprime. Since $ABC=2$(cube) we may assume $A=2i$(cube), $B=j$(cube), $C=k$(cube) where $i$, $j$, and $k$ are in the set $\{1,u,v\}$. Now all cubes in $\mathbb{Z}[u]$ are $0$ or $1$ mod $2$. Since $B+C$ is $0$ mod $2$, $j=k$. Since $ABC=2$(cube), $ijk$ is a cube and $i=j=k$. We may assume $i=j=k=1$. Then $A=2r^3$, $B=s^3$, $C=t^3$, and we may further assume that $s$ and $t$ are $1$ mod $2$. $s$ and $t$ are not both in $\{1,-1\}$ and it follows that $d$ is at least $\sqrt{27}$. Now look at $s+t$, $us+vt$ and $vs+ut$. They sum to $0$ and their product is $B+C=-2(r^3)$. They are congruent to $0$, $1$ and $1$ mod $2$, and the last 2 of them can't be equal since $s$ is not equal to $t$. Since each of them is at most $2d^{1/3}$, this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed. An almost identical argument gives what I think is the nicest proof of Fermat's Last Theorem for exponent 3.

  • $\begingroup$ Euler's proof seems to be a 2-descent rather than the 3-descent given above. I think the 3-descent argument is more elegant, and that undergrads will like it better. (I have used it in a number theory course). And I like introducing the Gaussian integers and Eisenstein integers towards the end of a course devoted to Z and Q as an easy introduction to algebraic number theory. $\endgroup$ May 28, 2011 at 16:08

There is a beautiful elementary way of solving the equation $$Ny^2-x^3=\pm1$$ when $N$ contains no prime factors that are of the form $6k+1$. There are only 7 solutions $(x,y,N)$ $$\{(2,3,1)(1,1,2)(23,78,2)(23,39,8)(2,1,9)(23,26,18)(23,18,72)\}$$ This was proven by J.H.E. Cohn in the article "The diophantine equations $x^3=Ny^2\pm 1$"

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    $\begingroup$ Alternatively you can reduce your equation to the cubic pell equation $x^3-2y^3=\pm 1$ and use $\mathbb Z [\omega]$, getting a solution similar to FLT for the exponent 3. A reference is for example E. Barbeau's "Pell equation". $\endgroup$ Sep 22, 2010 at 5:54
  • $\begingroup$ So the paper of J.H.E. Cohn contains solution to the Catalan Conjecture? $\endgroup$
    – Sergei
    Jun 5, 2015 at 5:32

Only the second proof below is new. In 1., I explain where the units in the cubic field are coming from, and 3. is a variation of Paul Monsky's proof.

  1. We try working in ${\mathbb Z}$ for as long as possible. Write $x^3 = y^2 - 1 = (y-1)(y+1)$; since the gcd of the factors on the right hand side divides $2$, there are two possibilities:

    a) $y$ is even; then $y+1 = \pm a^3$ and $y - 1 = \pm b^3$. Subtracting these equations gives $2 = a^3 - b^3 = (a-b)(a^2+ab+b^2)$, hence $a-b$ divides $2$. Going through all cases gives $(x,y) = (-1,0)$.

    b) $y$ is odd; then $y+1 = 2a^3$ and $y - 1 = 4b^3$. Subtracting these equations gives $1 = a^3 - 2b^3$.

    It remains to solve the last equation. One possibility is showing that the unit $1 - \sqrt[3]{2}$ is fundamental in the cubic number field ${\mathbb Q}(\sqrt[3]{2})$, and that its only powers of the form $a+b\sqrt[3]{2}$ have exponent $0$ or $1$.

    The other possibility is observing that this equation is a special case of the twisted Fermat cubic $x^3 + y^3 = 2z^3$. But Paul Monsky's proof shows directly that solving this equation is all that is needed.

  2. A more or less standard proof (this is basically a classical $2$-descent, but perhaps more direct than Euler's approach, which avoids number fields altogether - it is well known that on curves with a rational point of order $2$, a simple 2-decent can be performed by working within the rationals) proceeds as follows: write the equation in the form $$ y^2 = (x+1)(x+\rho)(x+\rho^2), $$ where $\rho$ is a primitive cube root of unity. The gcd of two factors divides $1-\rho$, hence there are two possibilities:

    a) $x+1 = \pm a^2$, $x + \rho = (-\rho)^e (a+b\rho)^2$, $x + \rho^2 = (-\rho^2)^e (a+b\rho^2)^2$. Since $\rho = (1+\rho)^2$ is a square, we can subsume the powers of $\rho$ into the square and find $x + \rho = \pm (a+b\rho)^2$ and $x + \rho^2 = \pm (a+b\rho^2)^2$. Subtracting these equations and dividing through by $\rho - \rho^2$ gives $1 = \pm b(2a-b)$, leading to $(x,y) = (0,\pm 1)$ and $(-1,0)$.

    b) Here we find $$ x+1 = \ \pm 3a^2, \quad x + \rho = \ \pm (1-\rho) (a+b\rho)^2, \quad x + \rho^2 = \ \pm (1-\rho^2) (a+b\rho^2)^2. $$ Adding the last two equations gives $2x-1 = \pm 3(a^2 - b^2)$. Eliminating $x$ from this and the first equation (where we actually have $x = 3a^2$ since $x \ge -1$) yields $(x,y) = (2,\pm 3)$.

  3. Let me give here my rendition of Paul's beautiful proof:

    Let $\alpha, \beta, \gamma \in {\mathbb Z}[\rho] \setminus \{0\}$. If $\alpha + \beta + \gamma = 0$ and $\alpha\beta\gamma = 2\mu^3$, then after a suitable permutation of the three numbers we have $\alpha = 0$ or $\beta = \gamma$.

    Proof. Let $(\alpha,\beta,\gamma)$ be a counterexample. Then $\alpha$, $\beta$ and $\gamma$ are pairwise coprime in ${\mathbb Z}[\rho]$, and after a suitable permutation we have $$ \alpha = 2 \rho^a A_1^3, \quad \beta = \rho^b B_1^3, \quad \gamma = \rho^c C_1^3. $$ Among all such counterexamples we now take one in which $N\alpha$ is minimal. Dividing all three numbers by $\rho^a$ we may assume that $a = 0$.

Cubes in ${\mathbb Z}[\rho]$ are $\equiv 0, 1 \bmod 2$ by Fermat's Little Theorem. Thus $0 \equiv \alpha \equiv \beta + \gamma \equiv \rho^b + \rho^c \bmod 2$, which implies $b = c$. Since $\alpha\beta\gamma$ is a cube, we must have $a = b = c = 0$.

Thus $$ \alpha = 2 A_1^3, \quad \beta = B_1^3, \quad \gamma = C_1^3. $$ Since $B_1^3 \equiv C_1^3 \equiv 1 \bmod 2$, we may assume that $B_1 \equiv C_1 \equiv 1 \bmod 2$ (after multiplying these numbers through by a suitable power of $\rho$).

Now set $\alpha_1 = B_1 + C_1$, $\beta_1 = \rho B_1 + \rho^2 C_1$ and $\gamma_1 = \rho^2 B_1 + \rho C_1$. Then

  • $\alpha_1 + \beta_1 + \gamma_1 = B_1(1+\rho+\rho^2) + C_1(1+\rho+\rho^2) = 0$.
  • $\alpha_1 \beta_1 \gamma_1 = B_1^3 + C_1^3 = \beta + \gamma = -\alpha = 2(-A_1)^3$.
  • $\beta_1 + \gamma_1 = (B_1 + C_1)(\rho+\rho^2) = - (B_1 + C_1) \ne 0$ since $\beta + \gamma = -\alpha \ne 0$.
  • $N(\alpha_1 \beta_1 \gamma_1) = N\alpha \mid N(\alpha\beta\gamma)$; if we had equality, it would follow that $N(\beta) = N(\gamma) = 1$, hence $\beta, \gamma = \pm 1$. But then $\beta = 1$, $\gamma = -1$ and $\alpha = 0$ against our assumptions.

Now descent finishes the proof: $(\alpha_1, \beta_1, \gamma_1)$ is another solution with $N(\alpha_1 \beta_1 \gamma_1) < N(\alpha\beta\gamma)$,

As a final remark I would like to point out that the article by Cohn in Gjergji Zaimi's answer uses "well known results" such as the solution of $x^4 - 2y^2 = \pm 1$ and $x^4 - 3y^2 = 1$. I do not know offhand how elementary the corresponding proofs are.

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    $\begingroup$ Well, $x^4-1=py^2$ implies one of $x^2\pm 1$ is a square, so the only non-trivial equation is $x^4+1=2y^2$. An elementary solution of the latter is not hard using the parametrization of the primitive solutions of $a^2+b^2=2c^2$. $\endgroup$ Jul 13, 2011 at 18:59
  • $\begingroup$ Franz, thank you for writing out these comments. If I could accept two answers, I would do so. $\endgroup$ Jul 20, 2011 at 19:53
  • $\begingroup$ Hi Franz, Can I ask in the second method, after we know gcd of any two factors divides $1-\rho$, how do we know the three factors take the form in your solution? $\endgroup$
    – JKDASF
    Apr 18, 2021 at 16:03
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    $\begingroup$ If the factors are coprime, they are squares up to units; if the gcd is $1-\rho$, the factors are products of $1-\rho$, a square and a unit. $\endgroup$ Apr 19, 2021 at 5:32
  • $\begingroup$ Dear Franz. Thank you for expanding "Paul's beautiful proof". Paul Monsky also mentioned that "An almost identical argument gives what I think is the nicest proof of Fermat's Last Theorem for exponent 3." I fully agree that this would be nice, but not sure how to prove the "b=c" part in this case. Can you help? $\endgroup$ Mar 3 at 12:43

More generally, $x^2-y^3=k$ is called the Mordell equation. For some values of $k$, you'll find elementary solutions in Mordell's Diophantine Equations, and in Uspensky and Heaslett's textbook.

EDIT: I've had a look at Mordell's book. He refers to his paper, The infinity of rational solutions of $y^2=x^3+k$, J London Math Soc 41 (1966) 523-525.


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