Hi laszloF,
as I am not completely satisfied with my above explanation, let me try another (perhaps better and more exact) approach:
The key to understand the effect of resistance change is the fact that two sources drive a current, but the current is allocated to one single source only. That's the only way to define something like an input resistance at one port. In principle, we have a similar situation by explaining the MILLER effect (however, with a resistance decrease). And here we have something like a bootstrap effect.
I'll try to explain the non-inverting case:
The signal voltage Vp (at the p-terminal of the input differential pair) causes a certain current Ip1 flowing into the terminal. However, at the same time there is another voltage Vn (the feedback voltage) at the inverting terminal of the differential pair. Now, because Vn causes also a current In=Ip2 through the p-terminal in a direction opposite to Ip1 (working principle of diff. stage), we have a total current Ip=Ip1-Ip2 through the p-input that, of course, is much more smaller than Ip1.
Because we allocate this current Ip to the signal voltage Vp only (although also the "source" Vn is partly responsible), this leads to a drastic increase of the effective input resistance at the p-terminal.
I am convinced that this is the most correct explanation.