How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# 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!

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.29.2" _i="3" _address="0.1.0.2.3"]

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.
[/et_pb_tab][et_pb_tab title="Food for Thought" _builder_version="3.29.2" hover_enabled="0" _i="6" _address="0.1.0.2.6"]
• 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.