 # 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.

• 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.

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