Pairwise independent affine hashing over a prime field

The integer 5 is prime: it is greater than one, and its possible positive proper divisors 2,3,4 do not divide it (remainders 1,2,1 respectively). Therefore F2 constructs the field used here, with residues 0,1,2,3,4. Its nonzero inverses are $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, $4^{-1}=4$, as $1,6,6,16$ all have remainder one. In particular division by the difference $3-1=2$ is valid.

Solving $a+b=2$ and $3a+b=4$ gives $2a=2$, hence $a=3\cdot 2=1$ and $b=1$. Direct substitution gives $h(1)=2$, $h(3)=4$. More generally outputs u,v give $a=3(v-u)$, $b=u-a$, a unique seed as in F1. Each joint output has probability $1\mathrm{/}25$ and each marginal $1\mathrm{/}5$, so these two values are independent uniform. This counts all slopes, including zero when u=v.

For every seed, $h(0)=b$, $h(1)=a+b$, and $h(2)=2a+b=2h(1)-h(0)$. Thus $h(0)=0$ and $h(1)=0$ force b=a=0 and consequently $h(2)=0$. The event $(h(0),h(1),h(2))=(0,0,1)$ has probability zero, whereas three independent uniform values would give $1\mathrm{/}125$. This is a concrete failure of three-wise independence, consistent with the pairwise conclusion.