# Does secure multiplication makes sense in the case of 2 inputs?

I am looking at a protocol to securely multiply numbers, so as to preserve privacy of inputs.

In particular, suppose there are three parties $$P_1$$, $$P_2$$ and $$P_3$$ and 2 numbers to multiply, $$a \in \mathbb{F}$$ and $$b \in \mathbb{F}$$ such that $$\mathbb{F}$$ is a finite field of prime order $$p$$. More precisely, $$a$$ is the $$P_1$$'s input, $$b$$ is $$P_2$$'s input and $$P_3$$ has no input. $$P_1$$ and $$P_2$$ shares their input among the parties using 2-out-of-3 secret sharing and, in this way, each party can compute a share of the product. Then, shares of the product are securely summed and all the parties can compute the result.

In this aim, I am wondering whether does it make sense when we have only two parties?

If $$p$$ is big enough (such that $$a \cdot b < p$$) , once $$P_1$$ knows $$a \cdot b \mod p$$, by knowing $$a$$, it could easily compute the secret value $$b$$.

For this reason, I do not see the point of using secure multiplication in this case. Could somebody help me understand this?

• Note that if $P_1$ picks $a=0$ they don't learn anything about $b$.
– SEJPM
Feb 29 '20 at 12:38
• True. I have seen the classical example of secure dating, in which the possible inputs were just 0 or 1. What about in general? Feb 29 '20 at 17:11