In this article, I will use Lagrange polynomial to solve some polynomial problems from Mathematical Olympiads.

**1) Lagrange’s interpolation polynomial**

**Theorem.** Let be a positive integer and are complex numbers such that if . Then there exists precisely one polynomial of degree not greater than such that .

*Proof.* If and are polynomials of degree not greater than such that then the polynomial of degree not greater than and has at least distinct roots, therefore is the zero polynomial and hence .

Now if then is a polynomial of degree not greater than and .

The polynomial is called Lagrange’s interpolation polynomial or Lagrange’s polynomial at nodes .

**Corollary.** If is a polynomial of degree not greater than and are complex numbers such that if then .

**2) Examples**

**Example 1.** Let and be polynomials. Find the remainder in the division of by .

*Solution 1.* Asume and are the quotient and remainder in the division of by , respectively. We have

Because and we have and , therefore use Lagrange’s interpolation polynomial at nodes and we have

*Solution 2.* We have and therefore

Because , we have is the remainder in the division of by .

**Example 2.** Let be a polynomial of degree satisfying Determine .

*Solution 1.* Use Lagrange’s interpolation polynomial at nodes we have

Therefore

*Solution 2.* We have the polynomial of degree and are its roots, therefore

for some constant .

Because , we have

therefore and

Hence .

Note that .

Therefore is equal to if is odd and equal to if is even.

**Example 3.** Let be a positive integer. Suppose are integers and Prove that at least one of the numbers where is greater than or equal to

*Solution.* Assume that . Use Lagrange’s interpolation polynomial at nodes we have

see the coefficient of we have

and therefore at least one of the numbers is greater than or equal to

**Example 4.** is a polynomial of degree such that

and Determine .

*Solution.* Use Lagrange’s interpolation polynomial at nodes we have

and therefore , but by hypothesis , hence we have , so .

**Example 5.** Prove that for any real number we have the following identity

*Solution.* Assume that and , then we need to prove

By Lagrange’s Interpolation polynomial at nodes we have

Now, in , setting we have

and we are done.

**Example 6.** Let and be real numbers satisfying

Determine .

*Solution 1.* Setting and

We have , and therefore by Lagrange’s Interpolation polynomial at nodes we obtain

From and we have

and hence

where

See coeficients of in both sides of we have and therefore

*Solution 2.* From hypothesis we have

and therefore and are roots of the polynomial

hence . By see coeficients of we have

so

**Example 7.** Let be a triangle with and . Prove that for any points in the plane we have

*Solution.* In the complex plane we assume that

By Lagrange’s Interpolation polynomial at nodes we have

see coeficients of in the both sides we obtain

and therefore

but and , therefore

and we are done.

**Example 8.** Find all polynomials with real coefficients such that for every positive integer there exists a rational with .

*Solution on AoPS.* Assume that is a polynomial satisfying for every positive integer there exists a rational with .

Clearly can’t be constant, so . For all take such that . gives using the Lagrange’s interpolation polynomial at nodes .

Then for some we have with leading coefficient . But then has rational root and also leading coefficient . So the denominator of divides , i.e. for all .

Now assume . Then for , and we find large enough with for all .

But contains exactly elements. So among the different numbers

we must find for some . This gives , a contradiction.

So we must have with , which is indeed a solution for the problem.

From Wiki: In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points and numbers the Lagrange polynomial is the polynomial of the least degree that at each point assumes the corresponding value (i.e. the functions coincide at each point). The interpolating polynomial of the least degree is unique, however, and it is therefore more appropriate to speak of “the Lagrange form” of that unique polynomial rather than “the Lagrange interpolation polynomial”, since the same polynomial can be arrived at through multiple methods. Although named after Joseph Louis Lagrange, who published it in 1795, it was first discovered in 1779 by Edward Waring and it is also an easy consequence of a formula published in 1783 by Leonhard Euler.

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