To do Algebraic K-theory, we need a technical condition that a ring $R$ satisfies $R^m=R^n$ if and only if $m=n$. I know some counterexamples for a ring $R$ satisfies $R=R^2$. 


Are there any some example that $R\neq R^3$ but $R^2 = R^4$ or something like that?


(c.f. if $R^2=R^4$, then we need that $R^3=R^5=\ldots =R^{2n+1}$ for any $n>1$)



Subtle counterexample to $m\neq n$ but $R^m=R^n$ for some ring $R$ ?
Rings that satisfy the condition 
$R^n \cong R^m \iff n=m$ are said to have *invariant basis number* or the *invariant basis property*. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$. 


Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that 


$$
R^m\cong R^{m'} \iff m=m'\ \mbox{or}\ m,m'\ge h \ \mbox{and} \ m\equiv m'\ (\mod k)
$$


A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$. 


There are earlier constructions of rings of type $(h,k)$ by Leavitt [W. G. Leavitt: 
Modules without invariant basis number, Proc. Am. Math. Sot. 8 (1957), 322-328], [W. G. Leavitt: 
The module type of a ring, Trans. Am. Math. Sot. 103 (1962), 113-130], but they are far more complicated. 


Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K\_0(R)\cong {\bf Z} \oplus \tilde{K}\_0(R)$. 


