# Can homomorphic operations on many different ciphertexts be multiplied by a same constant?

Assume Alice encrypts messages $m_1$,...,$m_n$ using secret keys $k_1$,...,$k_n$ on a homomorphic encryption scheme (BHHO), so she would get $c_1=Enc_{k_1}(m_1)$,...,$c_n=Enc_{k_n}(m_n)$. Then Alice sends a constant a , some random numbers $a_1$,...,$a_n$ and these ciphertexts to Bob. Can Bob do the following homomorphic operations? (The homomorphic encryption scheme requires that the ciphertext is multiplied by a random number) $$d_1=Enc_{k_1}(m_1)*{a_1}*a=Enc_{k_1}(m_1*{a_1}*a)$$ ... $$d_n=Enc_{k_n}(m_n)*{a_n}*a=Enc_{k_n}(m_n*{a_n}*a)$$

• What kind of homomorphic cryptosystem are we talking? Something like Paillier? – mikeazo Jan 11 '16 at 4:02
• Yes. It likes Paillier. – sam zhao Jan 11 '16 at 5:48
• BHHO (Crypto 2008) – sam zhao Jan 11 '16 at 6:30

Yes. Additive homomorphic encryption can also be used to obtain multiplication by a scalar. The way that this is done is by repeated addition, in exactly the same way as you do efficient exponentiation.

Recursively, define:

• If $a$ equals 1 then $mult(a,Enc_k(m)) = Enc_k(m)$

• If $a$ is even then $mult(a,Enc_k(m)) = mult\left(\frac{a}{2},Enc_k(m)\right)+mult\left(\frac{a}{2},Enc_k(m)\right)$

• If $a$ is odd then $mult(a,Enc_k(m)) = mult\left(\frac{a-1}{2},Enc_k(m)\right)+mult\left(\frac{a-1}{2},Enc_k(m)\right)+Enc_k(m)$