Are Euclid numbers squarefree? Are Euclid numbers squarefree ?
An Euclid number is by definition a Primorial number + 1.
See http://mathworld.wolfram.com/Primorial.html.
In notation the $n$ th Euclid number is written as $E_n = P_n+1.$
Thus I wonder about $a^2b = E_n$ for positive integer $a,b,n$ and $a>1$. 
I was thinking about Korselt's criterion : http://mathworld.wolfram.com/KorseltsCriterion.html
and Fermats little.
Im unaware of other techniques for proving the squarefree property apart from more elementary tools as gcd and basic modular aritmetic.
I cannot imagine infinite descent to work here ?
Maybe a binomium will help ?
I guess I missed something trivial and have a bad day.
Maybe $a^2b = E_n-2$ is easier ?
 A: Here's a heuristic argument of the kind that Hurkyl mentioned in a comment that gives a non-zero "probability" for all Euclid numbers to be squarefree.
The "probability" for a large number to be squarefree is
$$
\prod_{n=1}^\infty\left(1-\frac1{p_n^2}\right)=\frac6{\pi^2}\;,
$$
where $p_n$ is the $n$-th prime. The Euclid numbers grow so fast that we can asymptotically take them to be large numbers in the sense that they could potentially be divisible by the squares of all primes. (Any error incurred by this approximation will only cause us to underestimate their "probability" to be squarefree.) However, by construction $E_n$ is not divisible by the first $n$ primes. Thus, the $n$-th prime has $n-1$ chances of its square dividing a Euclid number. That makes the a priori "probability" of all Euclid numbers being squarefree  approximately
$$
\prod_{n=2}^\infty\left(1-\frac1{p_n^2}\right)^{n-1}\;.
$$
The first few terms of this product are to be taken with a large lump of salt, but we're not interested in them, only in the asymptotic behaviour and in the part left after Hagen's exploration up to $p_{45}=197$. Given that Hagen found all Euclid numbers up to $n=45$ to be squarefree, every prime now has $45$ fewer chances for its square to divide a Euclid number, so the a posteriori "probability" for all Euclid numbers to be squarefree is
$$
\prod_{n=47}^\infty\left(1-\frac1{p_n^2}\right)^{n-46}=\left(\frac{\pi^2}6\prod_{n=1}^{46}\left(1-\frac1{p_n^2}\right)\right)^{46}\prod_{n=47}^\infty\left(1-\frac1{p_n^2}\right)^n\;.
$$
The corrective factor before the infinite product is only about $1.035$ (computations 1 and 2), so we can focus on the infinite product and approximate it using asymptotic results for large $n$:
$$
\begin{align}
\prod_{n=47}^\infty\left(1-\frac1{p_n^2}\right)^n
&\approx\prod_{n=47}^\infty\left(1-\frac1{(n\log n)^2}\right)^n
\\
&\approx\prod_{n=47}^\infty\exp\left(-\frac1{n\log^2 n}\right)
\\
&\approx\exp\left(-\int_{47}^\infty\frac1{x\log^2 x}\mathrm dx\right)
\\
&=\exp\left(-\frac1{\log47}\right)
\\
&\approx0.77\;.
\end{align}
$$
Thus, together with the corrective factor of about $1.035$, there is now only roughly a $1$ in $5$ chance of finding a Euclid number that isn't squarefree even if there's no systematic reason preventing this. From a practical perspective, if there is one to be found, it's unlikely to be found by fully checking further Euclid numbers, which would soon become prohibitively difficult, since even going up to $n=100$ (with $p_n=541$) would only increase the product to $\exp(-1/\log100)\approx0.80$; it might be more efficient to check more numbers at higher $n$ for divisibility by smaller squares, which have a better hit rate.
Of course the same analysis holds for the numbers $P_n-1$.
[Update:]
Given the quote from Concrete Mathematics in Gerry's comment, we know that the square of no prime up to $p_{3000000}\approx5\cdot10^7$ divides a Euclid number; so the remaining chance of finding one that does is only approximately $1-\exp(-1/\log3000000)\approx6.5\%$.
