I always thought that part of the idea behind SSS was that there is only a single $t-1$ degree polynomial through $t$ unique points; that this holds when the polynomial coefficients are over the reals and also in particular for SSS when the elements are chosen from a finite field. However I'm running into (what I think is) a counterexample -- so I must be doing something wrong :)

Consider SSS with modulus is $4$, so we are working in $\text{GF}(4)$. Assume the secret to be shared is $S = 3$, and the polynomial chosen ends up being $f(x) = x^2 + 2x + 3 \mod 4$. We pick three points $(0,3), (1,2), (3,2)$ on $f(x)$ as the shares (I'm aware the y-intercept should not be chosen as a share in this type of SSS scheme, but it's not super relevant to the question at hand). When I attempt to reconstruct the secret by solving the system of equations for the polynomial coefficients defined by the shares, I get two possible polynomials for $f(x) = a_2 x^2 + a_1 x + S \mod 4$.

$S = 3 \mod 4$

$a_2 + a_1 + S = 2 \mod 4$

$9a_2 + 3a_1 + S = 2 \mod 4$

Solving this system for the coefficients yields $a_1 = 0, a_2 = 3$ and $a_1 = 2, a_2 = 1$ as solutions, implying that two degree $2$ polynomials go through the $3$ points $(0,3), (1,2), (3,2)$. Namely $g(x) = 3x^2 + 3 \mod 4$ and $h(x) = x^2 + 2x + 3 \mod 4$. I thought there should only be one degree $2$ polynomial through $3$ distinct points?

I think my understanding is wrong somewhere, and would appreciate anyone pointing it out!

  • $\begingroup$ Rereading this closely, maybe the mistake is that the integers $\mod 4$ are not a field, since $2$ doesn't have an inverse $\mod 4$.. so the integers $\mod 4$ are not isomorphic to $\text{GF}(4)$. Then, the error in my thinking is that the integers $\mod p^k$ for prime $p$ and $k \geq 1$ are isomomorphic to $\text{GF}(p^k)$. In this case, can someone clarify when the integers modulo $p^k$ are isomorphic to $\text{GF}(p^k)$? $\endgroup$
    – Gordon
    Commented Mar 24 at 16:20
  • $\begingroup$ For $k>1$, the integers modulo $p^k$ are not isomorphic to GF$(p^k)$; for $k=1$, they are. That's all there is to it. $\endgroup$ Commented Mar 25 at 16:25

1 Answer 1


You are operating in integers modulo $4$ which is not a field, it is $\mathbb{Z}_4.$ This means that there are zero divisors (for example $2\times2=0 \pmod 4$) and thus polynomial representation of functions is not unique.


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