# Cheating in (2,6)-Shamir Sharing Scheme

We consider a $$(2, 6)$$ Shamir secret sharing scheme over $$Z_{11}$$. Four players $$A,B,C,D$$ cooperate to find the secret, but one of them is cheating (so the share he claims to have might not be an actual share). They claim that their shares are;

$$A:(1,2)\quad B:(2,5)\quad C:(3,10)\quad D:(4,3)$$ Who is cheating and what is the secret?

I am using 4 equations:

\begin{aligned} 2 &= s + a & (1)\\ 5 &= s + 2a &(2)\\ 10 &= s + 3a &(3)\\ 3 &= s + 4a &(4) \end{aligned}

And solving for $$s$$ and $$a$$.

I don't seem to know the exact approach? How do I go about solving this problem?

• Hint: you need only two shares to recover the secret. What happens if you try the shares from AB, AC, AD? – fkraiem Apr 23 at 6:41

Also, remember that you will need to solve all those equations in $$\mathbb Z_{11}$$, i.e. using arithmetic modulo 11.