# Not Using RSA In OT1/2 Oblivious Transfer?

I was reading Oblivious Transfer (Wikipedia) and the 1-2 oblivious transfer. I was confused why RSA was needed. Why wouldn't this work?:

A:m1, m0

A:x1, x2 -> B

B:k, b

B:Xb+k -> v -> A

A:Xb+k-X1, Xb+k-X2 -> k1, k2

A:m0+k0, m1+k1 -> m'1, m'2 -> B

B:m'b-k -> mb

• @JohnHao: it works because (assuming $b=1$) the Bob receives $m'_1 = m_1 + (k^e + x_0 - x_1)^d$, but $k^e + x_0 - x_1$ is essentially a random number, and so Bob cannot recover the e'th root of it (this is the RSA problem). – poncho Dec 5 '19 at 14:42