Diophantine equation $x^3+x+a^2=y^2$ Prob: Show that for any positive integer $a$, Diophantine equation $$x^3+x+a^2=y^2$$ has at least one solution $(x, y)$, where $x, y$ are positive integers.
Source: My teacher.
Attempt:
First I tried $a=1$ and found the minimal solution $(x, y)=(72, 611)$, not a friendly one.
Now rewrite the equation as $$x(x^2+1)=(y-a)(y+a).$$ We hope that $x=b_{1}b_{2}, x^2+1=c_{1}c_{2}$ and $b_{2}c_{2}-b_{1}c_{1}=2a$. But how to determine these numbers? I got stuck here. One way promising is to relate to a Pell-type equation. 
Another thought is to set $x$ to be some polynomial like $2t^2$ so that the left side can be factorized furthermore. Still little progress.
Please help.
 A: Your first idea is a great idea! When working this out, I had slightly different variables, but it is equivalent to your starting.
Suppose positive integers $b, c, d, e$ satisfy $y+a=bc$, $y-a=de$, $x=bd$, $x^2+1=ce$, $bc-de=2a$, $ce-(bd)^2 = 1$.
I propose the following lemma:
The sequence $k_n$ is defined as follows: $k \in \mathbb{N}$, $k_0 = 0$, $k_1=1$, $k_{n+2} = kk_{n+1} + k_{n}$ for $n \geq 0$. Then $k_{n-1} k_{n+1} - k_{n}^2 = (-1)^{n} \quad \forall n \geq 1$.
Proof:
For $n=1$, this holds true. Now, assume true for $n$. Then for $n+1$, 
$$\begin{aligned}
k_{n} k_{n+2} - k_{n+1}^2 &= k_{n} (kk_{n+1} + k_{n}) - k_{n+1}^2\\
&= -k_{n+1} (k_{n+1} - kk_{n}) + k_{n}^2\\
&= -(k_{n-1} k_{n+1} - k_{n}^2)\\
&= -(-1)^n\\
&= (-1)^{n+1}
\end{aligned}$$
Thus, proven.
Now, if we let $c=k_3 = k^2+1$, $e=k_5=k^4+3k^2+1$, $bd=k_4 = k(k^2+2)$, then from the above lemma, we have $ce-(bd)^2=1$. Letting $d=2a, k=4a^2$, we get $b=2a(k^2+2)$ and $bc-de=2a(k^2+2)(k^2+1)-2a(k^4+3k^2+1) = 2a$. Thus, we have the result
$$x=bd=8a^2(8a^4+1)$$
$$y=4a(8a^4+1)(16a^4+1)-a$$
as a construction for all $a$, and does satisfy the solution you provided for $a=1$.
