By the second Bonse inequality we have $p_1p_2p_3\cdots p_m > p_{m+1}^3$ for $m\geq 5$, where $(p_m)_{m\geq 1}$ is the sequence $2,3,5,7,11,\ldots$ of the positive primes. Then of course $p_2p_3\cdots p_m > \dfrac {1}{2}p_{m+1}^3 > p_{m+1}^2$ for $m\geq 5$. A direct proof of this last assertion can be easily produced using Bertrand's postulate (a theorem in fact), a corollary of which states that $p_{m+1} < 2p_m$ for all $m\geq 1$. Now $p_2p_3p_4p_5 = 3\cdot 5\cdot 7\cdot 11 = 1155 > 169 = 13^2 = p_6^2$, thus the starting step for $m=5$ holds. Then $p_2p_3\cdots p_mp_{m+1} > p_{m+1}^2\cdot p_{m+1} > (2p_{m+1})^2 > p_{m+2}^2$, by the induction step. Back to our problem. For any odd prime $p$ with $p^2<n$ we must have $p\mid n$, otherwise $\gcd(p^2,n)=1$ and $p^2$ is not a prime. Say the largest odd prime $p$ with $p^2<n$ and $p\mid n$ is $p_m$, thus $p_3p_4\cdots p_m \mid n$, and so $p_3p_4\cdots p_m \leq n$. By the above result, for $m\geq 5$ we would have $n\geq p_3p_4\cdots p_m > p_{m+1}^2$, and $\gcd(p_{m+1}^2,n)=1$, but $p_{m+1}^2$ is not a prime. Therefore $m\leq 4$, and it is not difficult to see that $n=3\cdot 5\cdot 7 = \boxed{105}$ has all the properties, and is the largest such. This result was already obtained by Cseh László https://books.google.ro/books?id=DvX90EKMxGwC&pg=PA219&lpg=PA219&dq=Cseh+Bonse%27s+theorem&source=bl&ots=O7mbzhJS0t&sig=9Y4EdpNWjZ6gnUCU3VDgw8Zy0Oo&hl=ro&sa=X&ei=dFe7VJ6SFKSqywOg6oHgDg&ved=0CD4Q6AEwBA#v=onepage&q=Cseh%20Bonse's%20theorem&f=false. Bonse used his first inequality, stating that $p_1p_2p_3\cdots p_m > p_{m+1}^2$ for $m\geq 4$, to prove that $n=\boxed{30}$ is the largest positive integer such that any positive integer $1<k<n$ with $\gcd(k,n)=1$ is a prime. This has been featured in a very nice book, [Rademacher & Toeplitz - The Enjoyment of Math]. EDIT. Quite similar to this http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=621542&p=3715186#p3715186; same contest, earlier (not by much) year ...