Heuristic arguments due to (I believe) Delauney predict that every prime divides the order of the TateShafarevich group of infinitely many elliptic curves over $\mathbf{Q}$. However, is it even known that for every prime $p$, there's at least one elliptic curve over $\mathbf{Q}$ whose TateShafarevich group is finite and has order divisible by $p$?

2$\begingroup$ K. Matsuno, elliptic curves with large TateShafarevich groups (google will give you a pdf file) wrote in 2006 that the answer was not known. $\endgroup$ – Franz Lemmermeyer Mar 22 '11 at 18:45

3$\begingroup$ I believe that even if you replace ${\mathbb Q}$ by any fixed number field, this is still not known. (The papers by Matsuno and Alex Bartel (arxiv.org/abs/0805.1231) study Sha and Selmer growth in small extensions $K/{\mathbb Q}$, but the degree of $K$ grows with $p$.) $\endgroup$ – Tim Dokchitser Mar 22 '11 at 19:01

$\begingroup$ Wow! I somehow figured that this "must be known", but I guess I figured wrong... Thanks for the great references. $\endgroup$ – David Hansen Mar 22 '11 at 19:06

2$\begingroup$ Davidwhen the first edition of Silverman came out, there wasn't a single elliptic curve for which Sha was known to be finite! Facts about Sha are hard to come by... $\endgroup$ – Kevin Buzzard Mar 22 '11 at 19:35

2$\begingroup$ However, for every prime number $p$ and every integer $k$ there is an abelian variety $A/\mathbb{Q}$ such that the $p$torsion in the TateShafarevich group has $\mathbb{F}_p$dimension at least $k$. In this construction the dimension of $A$ grows with $p$. $\endgroup$ – Remke Kloosterman Mar 22 '11 at 20:25
Just to record an answer: yes, it is believed to be true that for each prime $p$ there is an elliptic curve $E_{/\mathbb{Q}}$ whose ShafarevichTate group contains an element of order $p$. But this is wide open.
One can however prove the following:
Theorem (R. Kloosterman): If you fix a prime $p$, then there is a number field $K_p$  i.e., depending on $p$  and an elliptic curve $E/K_p$ whose ShafarevichTate group contains an element of order $p$.
As Kloosterman remarked (in his 2005 paper on the subject, and also just now in the above comments) via a simple "Weil restriction" argument, this implies that there exists an abelian variety $A_{/\mathbb{Q}}$ whose ShafarevichTate group contains an element of order $p$. But now the dimension of $A$ goes to infinity with $p$.
Others have since worked out various improvements of Kloosterman's Theorem. Here are three different directions:
1) What is the minimal degree $d_p = [K_p:\mathbb{Q}]$ of a number field $K_p$ in above theorem? Kloosterman's argument gave $d_p = O(p^4)$. In a 2005 paper I showed that one can take $d_p = 2p^3$. In a 2010 paper with Shahed Sharif, we showed that one can take $d_p = p$. (Note that whether $d_p = p$ is in fact best possible is an open question. At one point I thought I had an argument to show that it was, but this was wrong. I now suspect that this is only best possible by a method proceeding along the lines of our construction. For instance, I believe it is conceivable that $d_p = 2$ for all $p$, and this has something to do with how Sha behaves in quadratic twists...)
2) What kind of elliptic curves $E$ can be used to produce these elements in Sha? Sharif and I showed that in fact one can start with any elliptic curve $E_0$ over $\mathbb{Q}$ and take its base change to $K_p$ a degree $p$ number field. (Similarly, one can start with any elliptic curve $E$ over any number field and get order $p$ elements of Sha in an extension field of degree $p$.) And in fact $p$ does not need to be a prime number here: it holds for every integer $n$. And in fact you can get as many elements of order $n > 1$ as you want.
3) What kind of control can one get over the field extension $K_p/\mathbb{Q}$? Can one for instance choose it to be Galois, abelian, or cyclic? Matsuno proves that for any cyclic degree $p$ extension $K_p/\mathbb{Q}$ one can find elliptic curves $E_{\mathbb{Q}}$ such that the base change to $K_p$ has as many order $p$ elements in its ShafarevichTate group as one wants. There are similar results along these lines (with dihedral extensions and elements of the Selmer group) by Alex Bartel.
P.S.: As Tim Dokchitser points out, replacing $\mathbb{Q}$ by any fixed number field does not help to answer the OP's question, although the results of Sharif and myself hold equally well in this relative setting. (I was going to ask "In what way could an arbitrary but fixed number field be any better than $\mathbb{Q}$?" but then I remembered an amazing result about $2$Selmer parity that only holds over certain imaginary number fields...due to Dokchitser and Dokchitser.)

2$\begingroup$  I attended a talk by Steve Donnelley in 2005, where he explained how one can get d_p<p. He applied a twodescent to a subclass of the examples of large pSelmer groups that Ed Schaefer and I constructed. At the time he did not work out a formula for $d_p$, but I figure that you get something like $d_p \sim 1/12 p$. Unfortunately, I never saw a written account of this. There is a further argument that shows that the examples of Ed and my are examples of big sha. But I did not write that up yet.  If you fix besides $p$ also $E$ (as you and Sharif did) then $d_p\geq p$ holds. $\endgroup$ – Remke Kloosterman Mar 23 '11 at 10:07