A number theory problem by Bogdan Simeonov

Mine is probably equivalent to Bogdan's, but I cheated a little.

Let $x = p/q$ with $p,q$ relatively prime. Write $p(p^2-q^2) = (aq)^3$ and note that the two factors on the left are relatively prime, so they're both cubes. Write $p = r^3$ . Write $b = aq/r$ . Then we get $r^6 - q^2 = b^3$ , so $q^2 + b^3 = r^6$ .

This is a well-studied equation. The solutions are

$(0,k^2,\pm k)$ : this leads to a denominator of $0$ , so forget this

$(\pm k^3, -k^2, 0)$ : this corresponds to $x = 0$ , so $a = 0$ .

$(\pm 3k^3, -2k^2, \pm k)$ : this gives $a = \pm 2/3$ , $x = \mp 1/3$ .

So the answer is $\fbox{4}$ .

Of course, this is a cheat because deriving the complete solution set to the Diophantine equation above requires arithmetic in ${\mathbb Z}[\sqrt{-3}]$ as in Bogdan's solution.

We will work in the field $Z(\sqrt{-3})$ , where the units are $\pm1,\pm{\varrho},\pm{\varrho^2}$ , $N(a+b\varrho)=a^2+ab+b^2$ .

$\textbf{Theorem}$ (You can skip this if you don't understand it): The equation $\xi^3+\eta^3=2\zeta^3$ has no solutions in the ring $Z[{\sqrt{-3}+1}/{2}]$ with $\xi\eta\zeta \neq 0$ , except $\xi=\eta=\zeta$

$Proof$ :We will consider the more general equation $\xi^3+\eta^3+2\epsilon\zeta^3=0$ , where epsilon is a unit.Let us call $N(\xi\eta\zeta)$ the $height$ of the solution $\xi,\eta,\zeta$ .The height is a positive number.A height of 1 is obtained when the numbers are $\pm1$ .Now let's assume the triple gives us the smallest height >1.Our three numbers are pairwise coprime.Now let us denote the numbers $\xi+\eta$ , $\varrho\xi+\varrho^2\eta$ and $\varrho^2\xi+\varrho\eta$ by $\alpha,\beta,\gamma$ in some order.Then

$\alpha+\beta+\gamma=0$ and $\alpha\beta\gamma=-2\epsilon\zeta^3$ .If $\delta$ is the gcd of alpha and beta, then $(\frac{\alpha}{\delta},\frac{\beta}{\delta})=1=(\frac{\alpha}{\delta},\frac{\gamma}{\delta})=(\frac{\gamma}{\delta},\frac{\beta}{\delta})$ .

Thus $\frac{\alpha}{\delta}=\epsilon_1.\xi_1^3$ , $\frac{\beta}{\delta}=\epsilon_2\eta_1^3$ , $\frac{\gamma}{\delta}=2\epsilon_3\zeta_1^3$ and $\frac{\zeta}{\delta}=\xi_1\eta_1\zeta_1$ , where $\xi_1,\eta_1,\zeta_1$ are pairwise coprime and the epsilons are units.Using the fact that $\alpha+\beta+\gamma=0$ and by multiplying by $\epsilon_1^{-1}$ , we arrive at

$\xi_1^3+\epsilon_4\eta_1^3+2\epsilon_5\zeta^3=0$ .Observing modulo 2, we would have that a cube is congruent to a unit.But that is only possible when that unit is $\pm1$ .But $\pm1$ is a cube.Thus we have that

$\xi_1^3+(\pm\eta_1)^3+2\epsilon_5\zeta^3=0$ .But that is an equation of the previous type.So we must have, by the minimality of the height of $\xi\eta\zeta$ , that either

$N(\xi_1\eta_1\zeta_1)=N(\frac{\zeta}{\delta}) \geq N(\xi\eta\zeta)$ or that $N(\xi_1\eta_1\zeta_1)=N(\frac{\zeta}{\delta})=1$ .In the first case we will have that $N(\xi\eta\delta)=1$ making all the numbers $\xi,\eta,\delta$ units, but then $\zeta$ must be a unit too,so our minimal height would be 1, contrary to our hypothesis that it is >1.

Now let's observe the second case.Since $\delta|\alpha-\varrho\beta,\delta|\alpha-\varrho^2\beta$ , then it must divide $(1-\varrho)\xi$ and $(1-\varrho)\eta$ .But $(\xi,\eta)=1$ , so delta is either a unit, or associated to the prime $1-\varrho$ .But from $N(\frac{\zeta}{\delta})=1$ , we have that $\zeta$ is associated with $\delta$ .But if $(1-\varrho)$ divides $\zeta$ , it must be that $(1-\varrho)^2$ divides $\zeta$ , contrary to them being associates(this follows, because the cubic residues mod $(1-\varrho)^4$ are -1,0 and 1).So we have that $\delta$ and $\zeta$ are units, from which it follows that $\alpha$ and $\beta$ are units.But then $(1-\varrho).\xi$ and $(1-\varrho).\eta$ are the sum of two units.But it is seen that this is only possible when $\xi$ and $\eta$ are units.But then our starting height must be 1, which is false.Thus the only solutions,where the three numbers are coprime and nonzero, are $1,1,1$ and $-1,-1,-1$ .Thus every nonzero solutions is obtained when the three numbers are equal.

$Q.E.D$

Now that we have finished proving this monstrosity, let's start with the problem (It starts now lol).Assume it is reducible.Then

$x^3-x-a^3=(x^2+\alpha.x+\beta)(x+\gamma)$

Comparing coefficients, we get that $\alpha=-\gamma$ , $\beta+\alpha\gamma=-1$ and $\gamma\beta=-a^3$ .Thus $\beta=\alpha^2-1$ and $\alpha\beta=a^3$ .So we get that $\alpha(\alpha^2-1)=a^3$ Now let's substitute $\alpha=\frac{m}{n}$ and $a=\frac{p}{q}$ .We get the equation $(m^3-mn^2).q^3=p^3n^3$ .But $(p,q)=1$ , so $q^3|n^3$ , leading to $q|n$ .If q is not equal to n, but smaller, there would exist a prime s, dividing n, but not q.The prime s would divide $m^3-mn^2$ , so $s|m$ , which is a contradiction, because $(m,n)=1$ .So $n=q$ and thus we have the equation $m^3-mn^2=p^3$ , where the triple is pairwise coprime.

$(m-n)m(m+n)=p^3$

Case 1:m and n are incongruent modulo 2.Thus the three factors of the LHS are coprime, so

$m-n=A^3,m=B^3,m+n=C^3$ .Solving this system, we get $2m=2B^3=A^3+C^3$ and $2n=C^3-A^3$ .But from the equation for 2m we would have that either B=0,A=-C, or A=B=C.the first case we would have that m=0, so $\alpha=0$ and a=0, which is a solution.In the second case, n=0, but that is impossible, since it is the denominator of $\alpha$ .

Case 2:m and n are both odd.

Here we would have that $(m-n,m,m+n)=(4A^3,B^3,2C^3)$ ,where A,B and C are coprime.WLOG $m-n=4A^3,m=B^3,m+n=2C^3$ .We can do this, because it doesn't matter if q is negative or positive.Solving this, we get that $m=2A^3+C^3=B^3$ and $n=C^3-2A^3$ .In the equation for have that $2A^3=B^3+(-C)^3$ .So we would have that A=B=-C, or that B=C,A=0.In the first case we would have that $A=B=\pm1$ and C is their opposite, because A,B and C are coprime.This tells us that $m=\pm1$ and $n=\mp3$ , which in turn gives that $\alpha=1/3,-1/3$ , so $a=2/3,-2/3$ .In the second case we would have that $B=C=\pm1$ , so $m=1$ and $n=\pm1$ , so $\alpha=\pm1$ , which means that a=0.

So all the solutions are $a=0,2/3,-2/3$ , where it is obvious that x=1/3 or -1/3 would be roots