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In Schnorr identification protocol, what happens if the prover uses r+c+x or rx+c.. etc. rather than r+cx?

Consider the Schnorr identification protocol. Let $x$ be secret key, $u$ be public key, $r(xleftarrow{$} mathbb{Z}_q )$ be random number with $u_r=g^r$that prover uses at first round, and $c(xleftarrow{$} C)$ be challenge that verifier challenges to the prover at second round. At the last round, the prover sends $s=r+cx$ to the verifier. Here is my::Listen

Consider the Schnorr identification protocol. Let $x$ be secret key, $u$ be public key, $r(xleftarrow{$} mathbb{Z}_q )$ be random number with $u_r=g^r$that prover uses at first round, and $c(xleftarrow{$} C)$ be challenge that verifier challenges to the prover at second round.
At the last round, the prover sends $s=r+cx$ to the verifier.

Here is my question. Why is $s$ of the form $r+cx$? Why not $r+c+x, rc+x, rx+c, rxc$?
When $s=rx+c$ or $rxc$, it seems that the verifier can not verify by computing since $g^{rx+c}=u^r g^c = u_r^x g^c $ and $g^{rxc}=u_r^{xc}=u^{rc}$ but the verifier does not know $r$ and $x$. However I don’t understand why the other do not work.

Mental health support prime for decentralization, say academics

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