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A question from Alibaba Global Mathematics Competition (number theory)

Prove $\displaystyle x^{3} +3y^{3} +9z^{3} -9xyz=1$ has infinitely many integer solutions.

The hint for the question is to transform the left side to a complex polynomial.

I found that:

Set $\lambda =e^{j2\pi /3}$ then the equation can be transformed to:

$$ \left( x+3^{1/3} y+3^{2/3} z\right)\left( x+3^{1/3} \lambda y+3^{2/3} \lambda ^{2} z\right)\left( x+3^{1/3} \lambda ^{2} y+3^{2/3} \lambda z\right)=1$$

but I don't know how to continue.

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    $\begingroup$ Is this an on-going contest? $\endgroup$ – Angina Seng Dec 23 '18 at 9:47
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    $\begingroup$ @LordSharktheUnknown It was held in Middle November. $\endgroup$ – Tianlalu Dec 23 '18 at 13:26
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We get a fairly clean appearance by introducing the matrix $$ M = \left( \begin{array}{ccc} 0& \beta &0 \\ 0&0& \beta \\ \beta &0&0 \end{array} \right) $$ with $$ \beta^3 = 3, $$ so that $$ M^3 = 3 I $$ Next we make a ring over integers with elements $$ xI + yM + z M^2 $$ so that your polynomial is $$ \det \left( xI + yM + z M^2\right) $$ If two elements of the ring have determinant one, so does their product. If the coefficients of both elements are positive, so are the coefficients of the product, and larger than either. We get a multiplication with elemnts binary quadratic forms, $$ \left( xI + yM + z M^2\right) \left( uI + vM + w M^2\right)= \left( (xu+3yw+3zv)I + (xv+yu+3zw)M + (xw+yv+zu) M^2\right) $$ So, if I have two triples $(x,y,z)$ and $(u,v,w)$ that evaluate to $1,$ so does this product triple, $$ \color{blue}{ (xu+3yw+3zv, xv+yu+3zw,xw+yv+zu) } $$ Beginning with the solution triple $(4,3,2)$ we may get a new larger positive solution from a given column vector $(x,y,z)^T$ by multiplying it by $$ W = \left( \begin{array}{ccc} 4&6 &9 \\ 3&4& 6 \\ 2 &3&4 \end{array} \right) $$ We get the sequence $$ (1,0,0), \; \; (4,3,2), \; \; (52,36,25), \; \;(649,450,312), \; \;(8104,5619,3896), \; \; \ldots $$

There are also solutions with negative elements, examples $$ (4,3,-4), \; (1,-18,12) $$ Indeed, we can simply begin with $(1,0,0)$ and keep multiplying by

$$ W^{-1} = \left( \begin{array}{ccc} -2&3 &0 \\ 0&-2& 3 \\ 1 &0&-2 \end{array} \right) $$

$$ (1,0,0), \; \; (-2,0,1), \; \; (4,3,-4), \; \;(1,-18,12), \; \;(-56,72,-23), \; \; \ldots $$

So far, the only solutions I get are the left hand column of some $W^n$ where $n$ is a positive or negative integer, or $0.$

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  • $\begingroup$ so many master here, i am lucky! $\endgroup$ – tangyao Dec 24 '18 at 15:26
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this is my anser

first suppose we have an answer x,y,z , for simplicity,

set $\displaystyle a=x,b=3^{1/3} y\ ,c=3^{2/3} z$

and take both sides pow3 ,we get

$\displaystyle ( a+b+c)^{3}\left( a+\lambda b+\lambda ^{2} c\right)^{3}\left( b+\lambda ^{2} b+\lambda c\right)^{3}$=1

$\displaystyle \begin{array}{{>{\displaystyle}l}} \left( a^{3} +b^{3} +c^{3} +3a^{2} b+3ab^{2} +3a^{2} c+3ac^{2} +3b^{2} c+3bc^{2} +6abc\right) *\\ \left( a^{3} +b^{3} +c^{3} +3\lambda a^{2} b+3\lambda ^{2} ab^{2} +3\lambda ^{2} a^{2} c+3\lambda ac^{2} +3\lambda b^{2} c+3\lambda ^{2} bc^{2} +6abc\right) *\\ \left( a^{3} +b^{3} +c^{3} +3\lambda ^{2} a^{2} b+3\lambda ab^{2} +3\lambda a^{2} c+3\lambda ^{2} ac^{2} +3\lambda ^{2} b^{2} c+3\lambda bc^{2} +6abc\right) \ =1 \end{array}$

$\displaystyle \begin{array}{{>{\displaystyle}l}} set\ A=a^{3} +b^{3} +c^{3} +6abc\ ,\ B=3a^{2} b+3b^{2} c+3ac^{2} \ ,C=3ab^{2} +3a^{2} c+3bc^{2} ,\ we\ get\\ ( A+B+C)\left( A+\lambda B+\lambda ^{2} C\right)\left( A+\lambda ^{2} B+\lambda C\right) =1\\ set\ A'=A\ ,\ \ B'=B/\left( 3^{1/3}\right) ,\ \ \ C’=C/\left( 3^{2/3}\right)\\ we\ get\\ \\ \ \ \left( A'+3^{1/3} B'+3^{2/3} C'\right)\left( A'+3^{1/3} \lambda B'+3^{2/3} \lambda C'\right)\left( A'+3^{1/3} \lambda ^{2} B'+3^{2/3} \lambda C'\right) =1\\ \\ then:\\ A'\ =a^{3} +b^{3} +c^{3} +6abc\ \ is\ interger\\ B'=B/\left( 3^{1/3}\right) =\left( 3a^{2} b+3b^{2} c+3ac^{2}\right) /3^{1/3} =3x^{2} y+9y^{2} z+9xz^{2} \ is\ interger\\ C'=\left( 3ab^{2} +3a^{2} c+3bc^{2}\right) /3^{2/3} =3xy^{2} +3x^{2} z+9yz^{2} \ is\ interger\\ \\ so\ we\ get\ \ A^{\prime 3} +3B^{\prime 3} +9C^{\prime 3} -9A'B'C'=1\ such\ that\ A' >a,B' >b,C' >c\\ we\ can\ repeat\ the\ process\ to\ get\ infinity\ solutions! \end{array}$

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First we find a non-trivial solution $(-2,0,1)$, ($(1,0,0)$ is trivial).

Claim: Let $\alpha=-2+0\cdot 3^{\frac13}+1\cdot3^{\frac23}$ and for any $n\in\Bbb N$, $$\alpha^n=x_n+y_n3^{\frac13}+z_n\cdot3^{\frac23},$$ then $(x_n,y_n,z_n)$ are solutions of the given equation.

Proof: The idea is similar to finding general solutions for Pell's equation, but with a little Galois theory (Pell's equation corresponds to a field extension with degree $2$, which must be a Galois extension).

The hint gives $$ \left( x+ay+a^2 z\right)\left( x+a \lambda y +a^2 \lambda ^{2} z\right)\left( x+a \lambda^2 y+a^2 \lambda z\right)=1$$ with $a=3^{1/3}$ and $\lambda =e^{j2\pi /3}$. Take both sides to the power of $n$.

Assume $(p,q,r)\in\Bbb Z^3$ is a solution, then $( p+aq+a^2 r)$, $( p+a \lambda q +a^2 \lambda^2 r)$ and $(p+a \lambda^2 q+a^2 \lambda r)$ are conjugate elements in the splitting field. Since $\Bbb Q(a,\lambda)/\Bbb Q$ is a Galois extension with degree $6$, any permutations between conjugate elements are isomorphic.

So $(p+aq+a^2 r)^n$, $(p+a \lambda q +a^2 \lambda^2 r)^n$ and $(p+a \lambda^2 q+a^2 \lambda r)^n$ are also conjugate of each other.

The first factor $(p+aq+a^2 r)^n$ is real so it corresponds to solutions.

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  • $\begingroup$ I have find the anser 1 minutes ago! $\endgroup$ – tangyao Dec 23 '18 at 13:26
  • $\begingroup$ so many master here, i am lucky! $\endgroup$ – tangyao Dec 24 '18 at 15:26
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This is a particular case of the so-called Mathews diophantine equation, 1889 (or sometimes "cubic Pell") $(M_{m,n})∶ x^3+my^3+m^2 z^3-3mxyz=n$, where $m,n \in \mathbf Z$. It can be put under a more tractable form by factorizing a certain « normic » expression in the ring $\mathbf Z[\mu,\omega]$, where $\omega$ is a primitive cubic root of $1$ and $\mu$ is the real cubic root of $m$. Precisely, the splitting field of $X^3-m$ is $\mathbf Q (\mu,\omega)$ whose Galois group is $S_3$, generated by $\tau: \mu \to \mu,\omega \to \omega^2$, and $\sigma: \omega \to \omega, \mu \to \omega\mu$. If 𝒩 is the norm (= product of conjugates) in $\mathbf Q(μ,ω)/\mathbf Q(\mu)$, then 𝒩$(x+yμ+zμ^2)=(x+yμ+zμ^2)(x+yωμ+zω^2 μ^2)(x+yω^2 μ+zωμ^2)=$ $x^3+my^3+m^2 z^3-3mxyz$ (using $1+ω+ω^2=0$), hence eq. $(M_{m,n})$ is simply equivalent to 𝒩$(x+yμ+zμ^2)=n$. The function 𝒩 being multiplicative, we are reduced to solving the case $(M_{m,p})$, where $p$ is a prime number, and the case $(M_{m,\pm 1})$.

Your problem belongs to the case to $(M_{m,\pm 1})$, with $m=3$. The equation 𝒩$(x+yμ+zμ^2)=\pm 1$ means simply that $x+yμ+zμ^2 \in \mathbf Z[\mu]$ is actually a unit. The ring of integers of $\mathbf Q (\mu)$ is classically known, see e.g. D. Marcus, "Number Fields", end of chap.2. In particular, for $m=3$, the ring of integers of $\mathbf Q (\mu)$ is $\mathbf Z[\mu]$ (which is a PID). We can apply Dirichlet's unit theorem, which states that the group of units has $\mathbf Z$-rank $1$. The fundamental unit is determined in exercise 37 (c) of Marcus, chap.5. Note that the resolution of the eq. $(M_{m,\pm 1})$ here is quite analogous to that of the quadratic Pell equation (cp. @Tianlalu).

NB. To my knowledge, a complete solution of $(M_{m,p})$ is not available. A natural approach would be parallel to that exposed in the book of D. Cox, "Primes of the form $x^2+ny^2$", but this requires deep results from CFT.

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  • $\begingroup$ so many master here, i am lucky! $\endgroup$ – tangyao Dec 24 '18 at 15:26

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