Since $x(x^2+1) = y^2-a^2=(y+a)(y-a)$, it is sufficient to show the existence of positive numbers $b,c,d,e$, which satisfy $y+a=bc, y-a=de, x=bd, x^2+1=ce, bc-de=2a, ce-(bd)^2=1$. We propose the following lemma: The sequence ${u_{n}}$ is defined as follows: $u\in N, u_{0} = 0, u_{1} = 1, u_{n+2}=uu_{n+1}+u_{n}$ where $n=0,1,2,3...$. Then $u_{n}u_{n+2} -u_{n+1}^2= (-1)^{n+1}, n \geq 0$. The proof of lemma: We prove by the induction on $n$. For $n=0$, it holds true. Assume it holds true for $n-1 (\geq 0)$. Then $u_{n}u_{n+2}-u_{n+1}^2 = u_{n}(uu_{n+1}+u_{n}) - u_{n+1}^2 = -u_{n+1}(u_{n+1}-uu_n)+u_n^2=-(u_{n+1}u_{n-1}-u_{n}^2) = (-1)^{n+1}$. Now, let $c=u_3=u^2+1, e=u_5 =u^4+3u^2+1, bd=u_4=u(u^2+2)$, then from the lemma we have $ce-(bd)^2=1$. Let $d=2a, u=4a^2$. Then $b=2a(u^2+2)$ and $bc-de=2a(u^2+2)(u^2+1)-2a(u^4+3u^2+1)=2a$. As a result, $$\begin{align*}x=bd=4a^2(16a^4+2) \\ y=2a(16a^4+2)(16a^4+1)-a\end{align*}$$ is one pair of positive integer solution of the equation.

Fun note: let $P=(0,a)$ be the obvious integral point on the elliptic curve $$y^2=x^3+x+a^2,$$ then $[3]P=(8a^2(8a^4+1),4a(8a^4+1)(16a^4+1)-a)$ is the integral point found above ($[3]P=P\oplus P\oplus P,$ where $\oplus$ is the addition on elliptic curves). This is also true for the elliptic curve $y^2=x^3+2x+a^2.$ In particular, let $P=(0,a)$ be the obvious integral point, then $[3]P=(4a^2(a^4+1),a(8a^8+12a^4+3))$ is another integral point on the curve.

My first PRKTST problem!!

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