# Is there an easy proof of this equation related to simultaneous Pell equations?

Working with the famous Baker-Davenport system of simultaneous Pell equations \begin{align} 3x^2-2 &= y^2, & 8x^2-7 &= z^2, \qquad(\star) \end{align} I am left, after a series of substitutions and reductions, with the equation $$u(u+1)\left(96u(u+1)+11\right) = v(v+1). \qquad(\dagger)$$ The point is now to prove that the only solutions with $u \ge 0$ are $u=0$ and $u=5$, yielding, respectively, the two known (and only) solutions $x=1$ and $x=11$ in ($\star$).

Are there any ways of attacking the formulation ($\dagger$), different from those which would naturally be applied to ($\star$), which might lead to a relatively straight-forward proof?

Thanks,
Kieren.

EDIT: For example, by unique factorization of integers, we can write $(v,v+1)=(abc,rst)$ for postive integers $a,b,c,r,s,t$ with $gcd(abc,rst)=gcd(v,v+1)=1$. Now, with appropriate permutation, we can write \begin{align} u &= ar, & u+1 &= bs, & 96u(u+1)+11=ct. \end{align} But all methods and manipulations I've tried from that point seem to lead to dead ends.

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Noting for documentation sake that OP posted, and then deleted, another question about these equation recently, mathoverflow.net/questions/143301/… – Gerry Myerson Sep 29 '13 at 5:36
@GerryMyerson: I found soon after I posted that the result was trivial, so I didn't think it worthy of remaining on MO. I did not know (1) that deleted posts are still visible, nor (2) that such documentation was required or expected. As you can tell from my reputation points and/or profile, I'm fairly new to MO and MSE — are these rules and expectations listed somewhere? Sorry, Kieren. – Kieren MacMillan Sep 29 '13 at 11:22
Deleted posts are visible to those with enough points. I don't know that documentation is required, but speaking for myself, I like people posting here to put all their cards on the table. – Gerry Myerson Sep 29 '13 at 12:18
@GerryMyerson: I'm certainly not trying to hide anything. I just didn't think that deleted observation was worth MO bandwidth, once I saw how trivial it was. – Kieren MacMillan Sep 29 '13 at 14:17

Sorry, it's probably no easier: both are looking for integer points on a genus-$1$ curve. In fact it's not hard to get from Baker-Davenport to your equation: multiplying the two equations in $(\star)$ yields $(3x^2-2)(8x^2-7) = (yz)^2$, and both $x$ and $yz$ must be odd, so we can write $(x,yz) = (2u+1,2v+1)$, and then subtracting $1$ and dividing by $4$ yields $(\dagger)$.
Yes, that's my "series of substitutions and reductions". I was (am) hoping that there would be (e.g.) a factorization approach applicable to ($\dagger$) that isn't applicable to ($\star$) directly. – Kieren MacMillan Sep 29 '13 at 11:25