Probabilistic RSA, IND CPA Suppose we have classic RSA scheme. Let $n=pq$, $e$ public, $d$ secret.

Then message(M), not empty(not 0), is encrypted by choosing random $r$ and computing $A=(r^e)^e$ and $B=(Mr)^e$

$(A,B)$ is then cipher text.

Why is this not IND-CPA secure? Have trouble grasping IND-CPA concept.. well concept is clear, but how to prove it?

Given above scheme, my thoughts are, since we are able to feed $M=1$ to oracle, we receive $r^{e^2}$ and $(r^2)^e$, but that thought leeds me absolutely nowhere.
I think i got it.

Query:

Let $m={1,2}$, and query $oracle(m)=(A,B)$

Checking:

Lets compute $s=B^e=(m^er^e)^e=m^{e^2}A$

Lets compute $sA^{-1}=\frac{m^{e^2}r^{e^2}}{r^{e^2}}=m^{e^2}$

if $s=1$ then $m=1$, $m\neq 1$ otherwise

Hopefully logic holds.