 # 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 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

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2014. Problem-23. You may use sequential hints to solve the problem.

## Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability from AMC 10A, 2010. Problem-23. You may use sequential hints to solve the problem.

## Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Triangle and Circle from AMC 10A, 2010. Problem-22. You may use sequential hints to solve the problem.