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
 A: 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 $a=b=c$.
A: Well not necessarily, You can make zero in lots of ways. For example: $5 - 3 - 2 = 12 - 7 - 5 = 0$. In this problem, $a=b=c$ because of the way that the coefficients have been arranged. You can't automatically state the the first terms are equal, because the end result is equal.
A: Write $P_1(k) = P_2(k)= P_3(k)=v$. Then 
$$
\pmatrix{ k^2 & -k & -1 \\ -1 & k^2 & -k \\ -k & -1 & k^2}
\pmatrix{ a \\ b \\ c}
=
\pmatrix{ v \\ v \\ v}
$$
and so
$$
\pmatrix{ a \\ b \\ c}
=
\frac{1}{k^6 - 4 k^3 - 1}
\pmatrix{k^4 - k & k^3 + 1 & 2 k^2 \\
2 k^2 & k^4 - k & k^3 + 1 \\
k^3 + 1 & 2 k^2 & k^4 - k}
\pmatrix{ v \\ v \\ v}
$$
Since the sums of the elements in each row are the same, we get $a=b=c$.
Alternatively, without any computation, just note that the first matrix is a circulant matrix and so has a circulant inverse.
On the other hand, we need $k^6 - 4 k^3 - 1 \ne 0$, that is, $k \ne (1\pm\sqrt5)/2$, the roots of $k^2-k-1$. This case needs to be handled separately but it still works. A necessary condition is $v=0$.
(Actually, $a=b=c\ne0$ implies $k^2-k-1=0$.)
A: From the expressions being zero we can not conclude that the coefficients are zero . We still need to show our work and prove that $a=b=c$ 
