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# Polynomial Functional Equation – Random Olympiad Problem

This beautiful application of Functional Equation is related to the concepts of Polynomials. Sequential hints are given to work out the problem and to revisit the concepts accordingly.

# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″ _i=”1″ _address=”0.0.0.1″]Find all the real Polynomials P(x) such that it satisfies the functional equation: $P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”3.29.2″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” hover_enabled=”0″ _i=”2″ _address=”0.1.0.2″][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″ _i=”0″ _address=”0.1.0.2.0″]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”3.29.2″ _i=”1″ _address=”0.1.0.2.1″]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?  [/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”3.29.2″ _i=”2″ _address=”0.1.0.2.2″]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!

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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 $P(x) = 0 / \frac{-1}{2}$.  [/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”3.29.2″ _i=”4″ _address=”0.1.0.2.4″]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.  [/et_pb_tab][et_pb_tab title=”Techniques Revisited” _builder_version=”3.29.2″ _i=”5″ _address=”0.1.0.2.5″]

• 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.
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• 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.