I'm studying about polynomials and interpolation with shamir secret sharing. I have some questions about it.
I found that there are always lower-degrees polynomials that intersects with 2 or more points (passing through 0) with another polynomial of degree at least 3 (in Zp).
Example: Supose a secret sharing scheme with threshold $5$, with the polynomial $x^4$ in $Z_{11}$ and the secret is $0$. Theoretically i could only recover the original polynomial and consequently the independent term (secret = $0$) if i have at least 5 points, right? However, if i combine correctly some shares $(x, f(x))$, i could recover another polynomial with the same independent term, and consequently, the secret. For example, instead of combining 5 points, i could combine the points $(1, 2, 8)$ with their evaluations and recover the polynomial $g(x) = 7x^2 + 5x + 0$. This gives to me the secret $0$.
Now supose all the polynomials $a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a0$ where $\{a_4, a_3, a_2, a_1, a_0\} \in Z_{11}$.
All polynomials of degree 3, 4, ..., $n$ have this property. Some secrets could be recovered with 2, 3, ... points and others only with 2 or 3 or more points. Note that even using less points (2 or 3) its not monotonic.
Questions:
1 - Is it correct to say that there is some kind of "failure" of shamir secret sharing scheme because of this combination of points?
2 - Could i prove that there are always less degree polynomials that intersects with high degree polynomials in point 0 and another one in $Z_p$?
Edit: I've made some experiments.
I have picked all combinations $C^p_2$ and $C^p_3$ of $(x, (fx))$ to all polynomials $ a_3x^3 + a_2x^2 + a_1x + a0$ where $\{a_3, a_2, a_1, a_0\} \in Z_{5}, Z_{7}$ and $a_3 > 0$. I can always recover the secret using less points than the threshold.
Thanks to everyone!