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The circuit satisfiability problem asks whether a Boolean circuit has an input on which its designated output is one. It is NP-complete.

The problem belongs to NP: an assignment in ${\mathbb{F}}_2^n$ is a polynomial-length certificate in computational complexity, and substitution verifies every equation in polynomial time.

For NP-hardness, reduce the Circuit satisfiability problem. Introduce one variable in ${\mathbb{F}}_2$ for every wire of a Boolean circuit. A logical negation gate $w=\neg u$ is enforced by $w=1-u$, and a logical conjunction gate $w=u\wedge v$ is enforced by $w=uv$. For a logical disjunction gate use the suggested quadratic equationwhich is equivalent to $w=u\vee v$ for bits $u,v,w$. Add the linear equation $w_{\mathrm{out}}=1$ for the designated output wire.

The construction introduces one variable and one equation per wire or gate, so it is a polynomial-time many-one reduction. A satisfying circuit input extends uniquely through its gates to a solution of the equations, and any solution gives a consistent accepting circuit computation. Since circuit satisfiability is NP-complete, quadratic-equation satisfiability over F2 is therefore

A language $L$ belongs to RP when a polynomial-time randomized algorithm rejects every $x\notin L$ and accepts every $x\in L$ with probability at least $1/2$.

Amplify the algorithm on length-$n$ inputs with $n+1$ independent repetitions, accepting if any repetition accepts. Its error on each positive input is at most $2^{-(n+1)}$, while it still never accepts a negative input. Choose all random bits for all repetitions in advance. By the union bound, the probability that this one fixed choice fails on at least one of the at most $2^n$ positive strings is at mostThus some random string works simultaneously for every input of length $n$. Hardwire that string into the polynomial-time computation and compile it into a Boolean circuit. The resulting polynomial-size circuit family decides $L$, proving