Prove: polynomial is uniquely defined by three of its values

[CynicusRex]

Homework Statement

Algebra - I.M. Gelfand, Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.
This means that if P(x) and Q(x) are polynomials of degree not exceeding 2 and P(x₁) = Q(x₁), P(x₂) = Q(x₂), P(x₃) = Q(x₃) for three different numbers x₁, x₂ and x₃, then the polynomials P(x) and Q(x) are equal.

Homework Equations

The Attempt at a Solution

The solution is simple (it's in the book) but I again approached it in a convoluted way in which I got stuck.
P(x) = ax²+bx+c, Q(x) = dx²+ex+f
P(x₁) = Q(x₁), P(x₂) = Q(x₂), P(x₃) = Q(x₃)

P(x₁) - Q(x₁) = 0
P(x₁) - Q(x₁) = P(x₂) - Q(x₂)
(P(x₁) - Q(x₁)) - (P(x₂) - Q(x₂)) = 0
(P(x₁) - Q(x₁)) - (P(x₂) - Q(x₂)) = P(x₃) - Q(x₃)
(P(x₁) - Q(x₁)) - (P(x₂) - Q(x₂)) - (P(x₃) - Q(x₃)) = 0
P(x₁) - Q(x₁) - P(x₂) + Q(x₂) - P(x₃) + Q(x₃) = 0
P(x₁) - P(x₂) - P(x₃) = Q(x₁) - Q(x₂) - Q(x₃)

$$(ax_{1}^{2}+bx_{1}+c)-(ax_{2}^{2}+bx_{2}+c)-(ax_{3}^{2}+bx_{3}+c)=(dx_{1}^{2}+ex_{1}+f)-(dx_{2}^{2}+ex_{2}+f)-(dx_{3}^{2}+ex_{3}+f)
$$$$ax_{1}^{2}+bx_{1}+c-ax_{2}^{2}-bx_{2}-c-ax_{3}^{2}-bx_{3}-c=dx_{1}^{2}+ex_{1}+f-dx_{2}^{2}-ex_{2}-f-dx_{3}^{2}-ex_{3}-f$$$$a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c=d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+e(x_{1}-x_{2}-x_{3})-f$$$$
a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c-d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})-e(x_{1}-x_{2}-x_{3})+f=0$$$$(a-d)(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+(b-e)(x_{1}-x_{2}-x_{3})-c+f=0$$Here I thought I was going to be able to prove you can only get 0 if a=d, b=e and c=f, but
$$(2-1)(4_{1}^{2}-3_{2}^{2}-2_{3}^{2})+((-1)-(-2))(4_{1}-3_{2}-2_{3})-8+6=0$$$$
(1)(16-9-4)+(1)(-1)-2=0$$$$
(1)(3)+(1)(-1)-2=0$$$$
3-1-2=0$$
I realize I'm making a mistake by plugging in numbers, because I'm changing the given property that P(x₃)=Q(x₃)
$$P(2)=2*2^{2}-1*2+8=14\neq Q(2)=2^{2}-2*2+6=6$$

Again, is it possible to prove it in a similary fashion as above? I understand the given solution below, but the simplicity didn't dawn on me :(

The solution:

Spoiler: Solution

Consider a polynomial R(x) = P(x) - Q(x). Its degree does not exceed 2. On the other hand, we know that R(x₁)=R(x₂)=R(x₃)=0;
in other words, x₁, x₂ and x₃ are roots of the polynomial R(x). But a polynomial of degree not exceeding 2, as we know, cannot have more than 2 roots, unless it is equal to zero. Therefore R(x) is equal to zero and P(x) = Q(x).

 

[PeroK]

There is a difference between mathematical "tricks" that are non-obvious insights that simplify a problem and good mathematical technique, which is standard and frequently repeatable, that simplifies a problem.

In this case, looking at the polynomial $P-Q$ is good technique and is something you must try to learn. It's not a one-off insight, it's a very common idea to tackle any problem where two things are equal.

PS note that you can also simply generalise the problem to polynomials of any order.

 

[epenguin]

I think proving 'a polynomial of degree not exceeding 2 is defined uniquely by three of its values' can be done easier.

There is a key point ppl often don't realize. We think there is linear algebra, and then there is polynomial algebra which is a different thing.

Whereas in fact a lot of polynomial algebra, most of what you can do, most of what you do do, is really linear algebra!

One is in the habit of thinking of a, b, c as given knowns and x as unknown; here you just have to think of a, b, c as 'unknowns' and you have plenty of 'knowns', x₁, x₁², P(x₁), x₂, ... and looking at it that way, the problem should be something very familiar.