# Is Paillier secure from known plaintext attack for single character message?

Assuming I have three messages m1,m2,m3 where m1=m2=m3=1 and I compute c1,c2 and c3.

Does that mean that c1=c2=c3 in cipher-text from ? If not, how many times can I encrypt a message m=1 and still produce a unique cipher-text ?

if not possible, can I solve the problem by putting Zeros as padding and still use the additive property ? (e.g. m1=01...m2=001 and m3=0001 and when I calculate x=c1+c2+c3 the decrypted result D{x}=3)

I just started working with cryptography, so please be understanding. :-)

Yes, Paillier encryption is secure from known plaintext attack (for single-character message, and any other supported message size). With high likelihood, three ciphertexts $$c_1$$, $$c_2$$ and $$c_3$$ for the same plaintext will be different.
When using public modulus $$n$$, each Paillier encryption draws a uniformly random number $$r$$ in range $$[0,n)$$ (some descriptions say $$[0,n^2)$$ but it turns out only $$r\bmod n$$ has an influence on the cryptogram; and in some descriptions it's added $$r$$ is coprime with $$n$$, but that's overwhelmingly likely for secure choice of $$n$$, thus can be omitted). Different $$r$$ will lead to different cryptograms $$c=g^m\cdot r^n\bmod n^2$$.
For proper implementation, by the birthday problem 101, the probability is about $$k^2/2n$$ that among $$k$$ ciphertexts for the same plaintext, any two are equal. For security against factoring, $$n$$ must have some thousand bits, thus said probability is entirely negligible for any feasible $$k$$ and secure choice of $$n$$. It is much more likely that whatever generates $$r$$ gets defective, and hickups, leading to identical ciphertexts.
Note: Pailler encryption is homomorphic only for messages that are straight integers $$m$$. Padding with zeroes on the left can't change $$m$$ if the homomorphic property must remain.