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Some days ago, I noticed that $3\cdot 5\cdot 7\cdot 11 +1=34^2$.
I am almost sure that if we denote four consecutive primes by $p, q, r, s$ then the equation

$$p\cdot q\cdot r\cdot s+1=x^2 \quad (1)$$

has finitely many solutions and here is the reason:
This equation is equivalent to $p\cdot q\cdot r\cdot s=(x-1)(x+1)$ which shows that $|p\cdot s-q\cdot r|=2$.
Now let $q=p+k, r=p+l, s=p+m$. We get $|p(m-k-l)-kl|=2$
which shows that $p\mid kl\pm2$ which means that $p\le kl+2$.
This should be a contradiction, because we expect that $k=q-p, l=r-p$ are not "too big" since they represent the gaps between primes who are consecutive and "almost" consecutive.
The problem is that we do not even have a proof for $k, l=\mathcal{O}(\sqrt p)$ which probably would be enough to finally solve the problem.

So, the fact that $p\le kl+2$ is a nice observation but does not lead anywhere.

Is there any way to show that equation (1) has finitely many solutions?

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  • $\begingroup$ Note that among $ p,q,r,s $ one or three numbers are of the form $ 4n+3 $ . $\endgroup$
    – Sylvain JULIEN
    Commented Mar 20, 2018 at 10:32
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    $\begingroup$ Also your conjecture would follow from the more general following one : the equation $ n-m=k $ with squarefree $ n $ and $ m $ and $ \omega(n)=\omega(m)=k $ has only finitely many solutions. $\endgroup$
    – Sylvain JULIEN
    Commented Mar 20, 2018 at 11:04
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    $\begingroup$ Equation (1) is also equivalent to the fraction $\frac{q}{s}$ having $\frac{(q-p)/2}{(s-r)/2}$ as a convergent and $\frac{p}{q}$ as a semiconvergent. This holds for the known solution: $\frac{5}{11}=[0;2,5]$ has convergent $\frac{1}{2}$ and semiconvergent $\frac{3}{7}$. This also means that the last coefficient in the continued fraction much be relatively large (larger than $\frac{s}{s-r}$). $\endgroup$
    – Max Alekseyev
    Commented Jul 25, 2019 at 0:27

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