Landweber exactness gives a criterion for when a complex oriented cohomology theory $E$ can be recovered from the formal group law over $E\_{\*}$ determined by the complex orientation. That is it gives a criterion for the following to hold (see for example [here](http://www.math.harvard.edu/~lurie/252xnotes/Lecture18.pdf))


$$E\_{\*}(X)\cong MU\_{\*}(X)\otimes\_{MU\_{\*}}E\_{\*}$$


Since the additive formal group law over $\mathbb{Z}$ is NOT Landweber exact, when $E=H\mathbb{Z}$ this isomorphism cannot hold for all spaces $X$.


My question is: what is an example of such a space?



Example of a space X exhibiting the Landweber non-exactness of the additive formal group over the integers?
Firstly, you ask for a space $X$. I will instead talk about finite spectra, but they become spaces if you suspend them enough times, so that does not really make a difference.


There is a kind of tautological answer to your question as follows. By the nilpotence technology of Hopkins, Devinatz and Smith, for suitable sequences of natural numbers $i\_k$ there are generalised Moore spectra $S/(v\_0^{i\_0},\dotsc,v\_n^{i\_n})$ such that 
$$ MU\_\*(S/(v\_0^{i\_0},\dotsc,v\_n^{i\_n})) = MU\_\*/(v\_0^{i\_0},\dotsc,v\_n^{i\_n}) $$
and there are cofibre sequences
$$ \Sigma^{|v\_n^{i\_n}|} S/(v\_0^{i\_0},\dotsc,v\_{n-1}^{i\_{n-1}}) \xrightarrow{} 
 S/(v\_0^{i\_0},\dotsc,v\_{n-1}^{i\_{n-1}}) \to 
 S/(v\_0^{i\_0},\dotsc,v\_n^{i\_n})
$$
with the obvious effect in $MU$-homology. Any spectrum of the form $S/(v\_0^{i\_0},v\_1^{i\_1})$ will be an answer for your question. Examples for this $n=1$ case were already constructed by Adams before the nilpotence theory was available (and this was a big part of the motivation for the nilpotence programme).


However, one might ask for a more elementary example. I think that one can proceed as follows. The Hopf map $\eta\colon S^1\to S^0$ has order $2$ and so extends to give a map $\eta'\colon S^1/2\to S^0$. Let $Q$ be the third suspension of the Spanier-Whitehead dual of the cofibre of $\eta'$. This has cells in dimensions $0$, $1$ and $3$. In mod $2$ homology, the bottom two cells are connected by $\text{Sq}^1$ and the top two are connected by $\text{Sq}^2$ so the cell diagram looks like a question mark and the complex is sometimes called the question mark complex. I think it works out that $MU\_\*Q=MU\_\*/(2,v\_1)x\oplus MU\_\*y$ with $|x|=0$ and $|y|=3$. On the other hand, we have $H\_\*Q=\mathbb{Z}/2x\oplus\mathbb{Z}z$ with $x$ mapping to $x$ and $y$ mapping to $2z$, so the map $\mathbb{Z}\otimes\_{MU\_\*}MU\_\*Q\to H\_\*Q$ is not surjective. However, the argument is a bit intricate. 


