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this problem is from my Gelfand's Algebra book.

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 $P(x_1)=Q(x_1), P(x_2)=Q(x_2), P(x_3)=Q(x_3)$ for three different numbers $x_1, x_2,$ and $x_3$, then the polynomials $P(x)$ and $Q(x)$ are equal.

I honestly don't know where to start.

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Hint: what are some roots of the polynomial $P-Q$? – Matthew Pressland Nov 6 '12 at 16:30
If there's $V(x)=P(x)-Q(x)$, and if $P(x_1)=Q(x_1), P(x_2)=Q(x_2), P(x_3)=Q(x_3)$, then $V(x_1)=V(x_2)=V(x_3)=0$, therefore $x_1,x_2,x_3$ are the roots of $V(x)$, but how can it be? if $V(x)$ is degree 2 polynomial. – Paul Dirac Nov 6 '12 at 16:38
The answer to the question "how can it be?" is the rest of the proof - see Hagen's answer. – Matthew Pressland Nov 6 '12 at 16:44
This fails if the coefficient ring is not a domain. e.g. the polynomials $\rm\:x^2\:$ and $1$ have the same values at the $4$ points $\pm1,\, \pm3$ over $\,\Bbb Z/8 =$ integers mod $8.\:$ So you need to specify the coefficient ring. – Bill Dubuque Nov 6 '12 at 16:45
I suspect the precalculus tag restricts the coefficient set... – copper.hat Nov 6 '12 at 16:59
up vote 3 down vote accepted

A polynomial of degree two is defined by three of it's values. Why? If we want to obtain a $2^{\rm nd}$ degree polynomial, we must specify three constants, $a_1,a_2,a_3$. But what are this constants? Say

$$p(x)=a_1x^2+a_2 x+a_3$$

Then, knowing three values, $b_1,b_2,b_3$ of points $x_1,x_2,x_3$, we have that

$$\begin{cases} a_1x_1^2+a_2 x_1+a_3=b_1\\ {}\\a_1x_2^2+a_2 x_2+a_3=b_2\\{}\\a_1x_3^2+a_2 x_3+a_3=b_3\end{cases}$$

and this is a system of three linear equations in three unknowns $a_1,a_2,a_3$, which we can always solve (meaning we can always determine its solutions, which might be none at all). This generalizes: an $n$ degree polynomial is uniquely determined by $n+1$ of its values.

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+1 The matrix formed from these equations is known as the Vandermonde matrix and the determinant has the form $(x_2-x_1)(x_3-x_1)(x_3-x_2)$ which is non-zero iff the $x_i$ are distinct. – copper.hat Nov 6 '12 at 16:55
@copper.hat Didn't know that, but seeing the book is titled "Algebra", it seemed the most natural approach. I'll take a look at that! – Pedro Tamaroff Nov 6 '12 at 17:03
It is very ill-conditioned from a numerical perspective, but shows 'invertibility' clearly. – copper.hat Nov 6 '12 at 17:05
Thank you Peter Tamaroff, but I don't understand how that proves that polynomial with degree not exceeding 2 is uniquely defined by 3 values. – Paul Dirac Nov 6 '12 at 17:24
@PaulDirac If you solve the linear system, you'll obtain what the coefficients of the polynomial are. Do you follow? – Pedro Tamaroff Nov 6 '12 at 17:29

Hint $\rm\:f(x) = ax^2\!+\!bx\!+\!c,\ a\ne 0\:$ and $\rm\:f(x) = f(y) = f(z)\:$ implies the following determinant $= 0$, having proportional first and last columns

$$\rm 0\, =\, \left | \begin{array}{ccc} \rm f(x) & \rm x & 1 \\ \rm f(y) & \rm y & 1 \\ \rm f(z) & \rm z & 1 \end{array} \right | \, =\, \left | \begin{array}{ccc} \rm ax^2 & \rm x & 1 \\ \rm ay^2 & \rm y & 1 \\ \rm az^2 & \rm z & 1 \end{array} \right | \, =\, a\,(x-y)(y - z)(z-x) $$

Thus, if the coefficient ring is a field (or domain), one of the RHS factors must be $0$. Since $\rm\:a\ne 0\:$ one of the other factors $= 0,\:$ i.e. the roots are not distinct.

Remark $ $ That the second determinant equals the first follows from the fact that the first columns are congruent modulo the others. More precisely, the second det arises by applying the elementary col operation $\rm\: col_1 \leftarrow col_1 - b\cdot col_2 - c\cdot col_3,\:$ yielding $\rm\: f(x) - b\cdot x - c\cdot 1 = ax^2,\:$ etc. The same argument works for higher degree polynomials. For the general case, it helps to know that the rhs matrix, after removing the factor of $\rm\:a,\:$ is the ubiquitous Vandermonde matrix,

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Thanks for help, I will add rep point once I reach 15 reputations. – Paul Dirac Nov 6 '12 at 17:55

Hint: $P-Q$ is a polynomial of degree not exceeding $2$ and has at least three roots $x_1, x_2, x_3$. How many roots would a nonzero polynomial of such degree have?

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Infinitely many, in general. But perhaps you meant to consider a certain more well-behaved class of polynomials? If so, then you should make that hypothesis explicit. – Bill Dubuque Nov 6 '12 at 16:48
Thank you Hagen for the hint, I understood that results, they contradict each other right? so does that mean that $P-Q$ is zero polynomial? Should I assume that? since polynomial of degree not exceeding 2 has 3 roots? – Paul Dirac Nov 6 '12 at 16:59
Thanks for help, I will add rep point once I reach 15 reputations. – Paul Dirac Nov 6 '12 at 17:56

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