<s>[INST] Three polynomials which have the same value for a variable 
Let $P_1(x)= ax^2-bx-c$, $P_2(x)=bx^2-cx-a$, and $P_3(x)= cx^2-ax-b$ be three quadratic polynomials, where $a,b$, and $c$ are non-zero real numbers. Suppose that there exists a real number $k$ such that $$ P_1(k) = P_2(k)= P_3(k) $$ Prove that $a=b=c$.

Hello everybody! The above is a question I got stuck on. This problem is from an Indian Olympiad. Here is what I did: 

After manipulating the given information, we derive at $(a-b)k^2-(b-c)k-(c-a)=0$ $(b-c)k^2-(c-a)k-(a-b)=0 $ and $(c-a)k^2 - (a-b)k - (b-c)$. Now the solution involves some addition and subtraction to factor it and get $a=b=c$.

But my question is: Can’t I just compare the coefficients of the three equations because they are all zero from which we get $b-c = c-a$, $c-a = a-b $ and $a-b = b-c$ and get $a=b=c$? If not, why?
Thanks
 [/INST] You can't do that because we need to show that this is the unique solution.
As an alternative, since the conditions


*

*$(a-b)k^2-(b-c)k-(c-a)=0$ 

*$(b-c)k^2-(c-a)k-(a-b)=0 $

*$(c-a)k^2 - (a-b)k - (b-c)=0$
are equivalent to the system $Mx=0$
$$\begin{bmatrix}A&-B&-C\\B&-C&-A\\C&-A&-B\end{bmatrix}\begin{bmatrix}k^2\\k\\1\end{bmatrix} =0$$
which, since $A+B+C=0$, is equivalent to
$$\begin{bmatrix}A&-B&-C\\B&-C&-A\\0&0&0\end{bmatrix}\begin{bmatrix}k^2\\k\\1\end{bmatrix} =0$$
Since $(k^2,k,1)$ and $(1,-1,-1)$ are linearly independent solutions we have that $\ker(M)\ge2$ and therefore
$$\det\begin{bmatrix}A&-B\\B&-C\end{bmatrix} =-AC+B^2=A^2+AB+B^2=(A-B)^2+3AB=0$$
and assuming wlog $a\ge b\ge c$ that is $A\ge B\ge 0$ we have that
$$(A-B)^2+3AB=0 \implies A=B=0 \implies C=0$$
that is $a=b=c$.
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