Solve $x^3-ax=by$ If $\gcd(x,y)=1$

Question

Solve the diophantine $x^3-ax=by$ If $\gcd(x,y)=1$.

Any hint? My first impression is $\gcd(x,b)>1$ and $x^2=ky+a$ for some integer $m$. I conclude that as long as there exists a square integer $u$ such that $ky+a=u^2$, we get proper solutions each and every time. Is my approach correct? because I was thinking that at least one of the solutions $x$ must have a common prime factor with $y$ Since the constant term is a multiple of $y$.Please help.

Answer 1

Hint:

Since $\gcd(x,y)=1$ therefore there exists $m_0,n_0 \in \mathbb{Z}$ such that $$m_0x+n_0y=1.$$ In fact for any $r \in \mathbb{Z}$ we can say that $$(m_0-ry)x+(n_0+rx)y=1,$$ and these are the totality of solutions.

From this we get $$[(m_0-ry)x^3]x+[(n_0+rx)x^3]y=x^3.$$ This can be rewritten as $$x^3-\color{red}{[(m_0-ry)x^3]}x=\color{red}{[(n_0+rx)x^3]}y.$$

Answer 2

If $gcd(x,y)=1$, then $x|b$. For any divisor $d$ of $b$, if $x=d$, then $y=\frac{d^2-a}{b/d}$ must be an integer. This depends on $a$ and $b$ and chosen $d$. There is always the solution $x=b$ and $y=b^2-a$ if $gcd(b, b^2-a)=1$. For instance, $x^3-3x=3y$ does not have solutions with $gcd(x,y)=1$.