Problem of Integer Pairs

Question

Prove that there exists infinitely many pairs of positive integers $(m,n)$ satisfying the following properties:

1. $\gcd(m,n)=1.$

2. $(x+m)^3=nx$ has three distinct integer solutions.

Answer 1

If $p\mid n$ and $p\mid x$ then $p\mid m$ hence gcd$(n,x)=1$ and both $n$ and $x$ are cubes.

Let $n=a^3,x=b^3$ then $x+m=ab, m=ab-x=ab-b^3.$

gcd$(m,n)$=gcd$(ab-b^3,a^3)=1$, we get gcd$(a,b)=1$.

$(x+m)^3-nx=(x-b^3)(-a^3 + 3 a^2 b^2 - 3 a b^4 + b^6 + 3 a b x - 2 b^3 x + x^2)$

Hence the equation $-a^3 + 3 a^2 b^2 - 3 a b^4 + b^6 + 3 a b x - 2 b^3 x + x^2=0$ has two distinct integer solutions, $\Delta =(-2 b^3 + 3 a b)^2 - 4 (-a^3 + 3 a^2 b^2 - 3 a b^4 + b^6)=a^2 (4 a - 3 b^2)$ must be a square number.

Let $4a-3b^2=t^2,a=\dfrac{t^2+3b^2}4,t\equiv b \pmod 2.$

For example, let $t=2u,b=2v$, then $a=u^2 + 3 v^2,m= v (u - v) (u + v),n=(u^2 + 3 v^2)^3,$ and $$(x + m)^3 - n x=(u^3 - 3 u^2 v + 3 u v^2 - v^3 - x) (8 v^3 - x) (u^3 + 3 u^2 v + 3 u v^2 + v^3 + x).\tag 1$$

1.gcd$(m,n)=1$ iff gcd$(a,b)=1$ iff gcd$(u,v)=1$ and $u\not \equiv v\pmod 2.$

2.Equation (1) has distinct integer solutions except for finitely many $u,v.$

Now you can pick $u,v$ that satisfying those properties.