In his BAII, Jacobson gives the following exercise (Exercise 3 in §4.2 of Basic Algebra II), which he attributes to McCrimmon.
Let $A$ be an (associative) ring with unity. Let $\mathfrak{R}\left(A\right)$ denote the Jacobson radical of $A$.
Show that $z\in\mathfrak R(A)$ iff for each $a\in A$ there exists $w\in A$ such that $z+w=zaw=waz$.
I have found this reference which is authored by McCrimmon and says "Communicated by Nathan Jacobson", but honestly, I cannot make much out of it (for example, I don't know what an homotope of an (alternative?) algebra is).
I have tried some algebraic manipulations using that whenver $z$ is in the radical, every element of $Az$ is left quasi-invertible and every element of $zA$ is right quasi invertible: suppose $(1-w)(1-az)=1$ and $(1-za)(1-w')=1$ that is $$\begin{eqnarray}\tag 1 w+az=waz\\ \tag 2 w'+za=zaw'\end{eqnarray}$$
Multiplying the first on the right by $aw'$ and the second on the left by $wa$ and subtracting gives $azaw'=waza$ and multiplication of the first ones by $a$ to the left and to the right resp gives $waza=wa+aza$ and $azaw'=aw'+aza$ so cancelling gives $wa=aw'$. Then we get $zw=w'z$ by the same token. This gives finally that $w+az=waz$ can be turned into $w+az=azw$, so $(1-az)(1-w)=1$ too. The same with the other. This proves that $1-az$ and $1-za$ are both q.i. Can anyone help further?
Suppose first that $z \in A$ is such that for all $a\in A$, there is $w \in A$ satisfying $z+w=zaw=waz$.
Then from $z+w=waz$ you can get $$z+w-waz=0\\ z+w(1-az)=0\\ az+aw(1-az)=0\\ 1-az-aw(1-az)=1\\(1-aw)(1-az)=1 $$
This shows $1-az$ is left-invertible for any $a$. Similarly, $z+w=zaw$ shows that $1-az$ is right-invertible for any $a$. Hence, $1-az$ is invertible for any $a$, so $z$ is in the radical.
Supposing now that $z$ is in the radical and $a\in A$, we seek a $w \in A$ such that $z+w=zaw=waz$. Solving $z+w=zaw$ for $w$ you get $$w-zaw=-z\\(1-za)w=-z\\w=-(1-za)^{-1}z$$. The inverse exists, of course, since $za$ is (left and right) quasiregular.
From our computation above, we already know $z+w=zaw$, so all that's left is to confirm that $zaw=waz=z+w$.
We have $$waz=-(1-za)^{-1}zaz\\=-(1-za)^{-1}z+(1-za)^{-1}z-(1-za)^{-1}zaz\\ =-(1-za)^{-1}z+(1-za)^{-1}(1-za)z\\=-(1-za)^{-1}z+z=w+z$$