Select Page

# Understand the problem

Find all the real Polynomials P(x) such that it satisfies the functional equation:

$P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.

Unknown

##### Topic
Functional Equation, Polynomials
7/10
##### Suggested Book
Excursion in Mathematics

Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

Well, it is really good that the information polynomial is given! You should use that.

What is the first thing that you check in a Polynomial Identity?

Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $n^2$. But did you observe something fishy?

Now rewrite the equation as $P(2P(x)) - 2P(P(x)) = P(x)^{2}$.

Do the Degree trick now… You see it right?

Yes, on the left it is $n^2$ and on the RHS it is $2n$.

So, there are two cases now… Figure them out!

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function.

Case 2: $P(2P(x)) - 2P(P(x))$ has coefficient zero till $x^2n$.

We will study case 1 now.

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function.

$P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $P(2y) - 2P(y) = y^{2}$ where $y = P(x)$.

Now, expand using $P(x) = ax^2 + bx +c$, it gives $2ay^2 -c = y^2$… Now find out all such polynomials satisfying this property.

For e.g. $\frac{x^2}{2}$ is a solution.

If P(x) is constant, prove that $P(x) = 0 / \frac{-1}{2}$.

Case 2: $P(2y) - 2P(y) = y^{2}$.

Assume a general form of P(x) = $latex$and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.

Now, we have already solved it for quadratic or less degree.

• Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
• Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
• Find all polynomials $P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
• Find all polynomials $$P(cP(x)) – d P(P(x)) = P(x)^{2}$$ depending on the values of c and d.

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## RMO 2019 (Maharashtra Goa) Adding GCDs

Can you add GCDs? This problem from RMO 2019 (Maharashtra region) has a beautiful solution. We also give some bonus questions for you to try.

## Number Theory, Ireland MO 2018, Problem 9

This problem in number theory is an elegant applications of the ideas of quadratic and cubic residues of a number. Try with our sequential hints.

## Number Theory, France IMO TST 2012, Problem 3

This problem is an advanced number theory problem using the ideas of lifting the exponents. Try with our sequential hints.

## Algebra, Austria MO 2016, Problem 4

This algebra problem is an elegant application of culminating the ideas of polynomials to give a simple proof of an inequality. Try with our sequential hints.

## Number Theory, Cyprus IMO TST 2018, Problem 1

This problem is a beautiful and simple application of the ideas of inequality and bounds in number theory. Try with our sequential hints.

## Number Theory, South Africa 2019, Problem 6

This problem in number theory is an elegant applciations of the modulo technique used in the diophantine equations. Try with our sequential hints

## Number Theory, Korea Junior MO 2015, Problem 7

This problem in number theory is an elegant application of the ideas of the proof of infinitude of primes from Korea. Try with our sequential hints.

## Inequality, Israel MO 2018, Problem 3

This problem is a basic application of triangle inequality along with getting to manipulate the modulus function efficently. Try with our sequential hints.

## Number Theory, Greece MO 2019, Problem 3

This problem is a beautiful application of prime factorization theorem, and reveal how important it is. Try with our sequential hints.

## Algebra, Germany MO 2019, Problem 6

This problem is a beautiful application of algebraic manipulations, ideas of symmetry, and vieta’s formula in polynomials. Try with our sequential hints.