To any given positive integer $m$ there exists a positive integer $x,$ such that $x^3+x+m^2$ is a square Let $m$ be a positive integer. Show that
there exists positive integer $x$ such that
$x^3+x+m^2$ is a square.
This problem I can't have any idea,How to prove it?
 A: This problem is about finding certain integral points on the elliptic curve
$$E:y^2=x^3+x+m^2.$$
The usual approach is to find rational points first, and pick up integral points from them.
As for the rational points, the group law of points on $E$ will help us to generate more rational points from given ones, and there are explicit formulas to do that.
In this answer, I use the geometric interpolation instead of the addition formulas to provide more intuition. The important fact is that, given a straight line passing through two points on $E$, it will intersect with $E$ at a third point in general. And if the first two points are both rational, so is the third one (Note that, with the given points, there is no need to solve cubic equation for the third one, but use Vieta's formula to express it in term of the first two. This observation derives the rationality). Moreover, this also holds in the limit situation when the first two points coincide and the line become the tangent of $E$ at that point.
Here we start the calculation. In the beginning, we have the obvious point $P=(0,m)\in E(\mathbb{Q})$. Although we also have $-P=(0,-m)\in E(\mathbb{Q})$, but the line connect $P$ and $-P$ produce nothing more. So we have to consider the tangent at $P$.
Denote $F(x,y)=y^2-x^3-x-m^2$, then the tangent at $P$ is given by
$$0=\frac{\partial F}{\partial x}\biggr|_P(x-x(P))+\frac{\partial F}{\partial y}\biggr|_P(y-y(P))=-x+2my-2m^2.$$
Joint it with equation of $E$ to get the third intersection, it will be $-2P=(1/(4m^2),1/(8m^3)+m)$. It is rational as expected but not integral, so we should move on.
Then we compute the line passing through $-2P$ and $-P$
$$(y(-2P)-y(-P))(x-x(-P))=(x(-2P)-x(-P))(y-y(-P)).$$
It will intersect with $E$ at the third point $3P=(8m^2+64m^6,512m^9+96m^5+3m)$. That is what we need for the problem.
This approach seems to be systematic but actually can't guarantee to work. In fact, it is surprising to me that $3P$ could be integral in this problem, as $nP$ tends to have much bigger denominator for bigger $n$.  

EDIT:
For the fact that $3P$ is integral while $2P$ is not, I think it is related to the different behavior of division polynomials with even and odd index.
Given a point $P$ on the elliptic curve $y^2=x^3+Ax+B$, its multiple $mP$ can be expressed as
$$mP=\biggl(\frac{\phi_m(P)}{\psi_m(P)^2},\frac{\omega_m(P)}{\psi_m(P)^3}\biggr)$$
for certain sequences of polynomials $\psi_m,\phi_m,\omega_m\in\mathbb{Z}[A,B,x,y]$. The first four terms of $\psi_m$ are
\begin{align}
\psi_1&=1,\\
\psi_2&=2y,\\
\psi_3&=3x^4+6Ax^2+12Bx-A^2,\\
\psi_4&=4y(x^6+5Ax^4+20Bx^3-5A^2x^2-4ABx-8B^2-A^3).\\
\end{align}
It can be seen and proved that the odd terms belong to $\mathbb{Z}[A,B,x]$, while the even terms belong to $y\cdot\mathbb{Z}[A,B,x]$.
Back to the question, we have $A=1,x(P)=0$ which makes $\psi_3(P)=-1$ much smaller than usual. On the other hand, $y=m$ makes $\psi_2(P)=2m$ can't be that small. From this point of view, the result looks reasonable.
A: The choice
$$x=64m^6+8m^2$$
works.
Then
$$x^3+x+m^2=(512m^9+96m^5+3m)^2.$$
