It is simple manupulation, equating the terms and finding out the values of b,c and d, then, solving the quadratic.

Set $P(x)=0 \Rightarrow Q(0)=0 \Rightarrow d=0$. Expanding both sides gives: $$c=\frac{1}{2}$$$$b=d-\frac{1}{2}=-\frac{1}{2}$$ Putting these values in $P(x)$, gives $-1$ and $\frac{1}{2}$ as it's roots. $$P(Q(x))=0\Rightarrow Q(x)=-1 \text{ or } \frac{1}{2}$$$Q(x)=-1$ gives imaginary roots and $Q(x)=\frac{1}{2}$ gives $x=\frac{1}{2}\text{ and } -1$.

rd1452002 wrote: Set $P(x)=0 \Rightarrow Q(0)=0 \Rightarrow d=0$. Expanding both sides gives: $$c=\frac{1}{2}$$$$b=d-\frac{1}{2}=-\frac{1}{2}$$ Putting these values in $P(x)$, gives $-1$ and $\frac{1}{2}$ as it's roots. $$P(Q(x))=0\Rightarrow Q(x)=-1 \text{ or } \frac{1}{2}$$$Q(x)=-1$ gives imaginary roots and $Q(x)=\frac{1}{2}$ gives $x=\frac{1}{2}\text{ and } -1$. How can you set p(x) =0?

rd1452002 wrote: Set $P(x)=0 \Rightarrow Q(0)=0 \Rightarrow d=0$. Expanding both sides gives: $$c=\frac{1}{2}$$$$b=d-\frac{1}{2}=-\frac{1}{2}$$ Putting these values in $P(x)$, gives $-1$ and $\frac{1}{2}$ as it's roots. $$P(Q(x))=0\Rightarrow Q(x)=-1 \text{ or } \frac{1}{2}$$$Q(x)=-1$ gives imaginary roots and $Q(x)=\frac{1}{2}$ gives $x=\frac{1}{2}\text{ and } -1$. That's only if you have a $x$ for which $P(x)=0$. I got to this point too, assuming we have a root (for which $P(x)=0$, because if there is no $x$ satisfying this, then obv. there is no root for $P(Q(x))$ either) but I went in this direction because I thought in the end we'll get that there won't be an actual root, but x=1/2 and -1 are indeed ones. You showed that if it has roots, there is two. But if there are no roots, then obv. there are no roots, but will there be satisfying functions in both cases? I highly doubt so. Guess there'll be only one P(x) and Q(x) solution tbh. And only one solution could satisfy one of the two conditions: has roots ($x=1/2$ and $-1$) or no roots at all..

misinnyo wrote: That's only if you have a $x$ for which $P(x)=0$. I got to this point too, assuming we have a root (for which $P(x)=0$, because if there is no $x$ satisfying this, then obv. there is no root for $P(Q(x))$ either) but I went in this direction because I thought in the end we'll get that there won't be an actual root, but x=1/2 and -1 are indeed ones. You showed that if it has roots, there is two. But if there are no roots, then obv. there are no roots, but will there be satisfying functions in both cases? I highly doubt so. Guess there'll be only one P(x) and Q(x) solution tbh. In the paper I didn't actually set $P(x)=0$, but did complete expansion, got $d=b+1/2$, $b^2+(c-d)b +d=0$ which gives $d=0$ and $b=-1/2$.

NOLF wrote: How can you set p(x) =0? You can. Since $P(x)Q(x) = Q(P(x))$ for all real $x$, and $P(x)Q(x)$ and $Q(P(x))$ are polynomials, so the polynomials themselves must be identical. Now we can safely plug in any complex $x$ too.

Can we set $ Q(x)=x$? Then we get either x=1 or P(x)=0 => $Q(0)=0$

Giffunk wrote: Can we set $ Q(x)=x$? Then we get either x=1 or P(x)=0 $Q(x) = x$ is true for only finitely many $x$, so it is the case that $x=1$ or $P(x) = 0$ for those $x$ that satisfy $Q(x) = x$. Besides, I do not see how you are getting that from the equation.

biomathematics wrote: Giffunk wrote: Can we set $ Q(x)=x$? Then we get either x=1 or P(x)=0 $Q(x) = x$ is true for only finitely many $x$, so it is the case that $x=1$ or $P(x) = 0$ for those $x$ that satisfy $Q(x) = x$. But there are only finitely many $x$

biomathematics wrote: Giffunk wrote: Can we set $ Q(x)=x$? Then we get either x=1 or P(x)=0 $Q(x) = x$ is true for only finitely many $x$, so it is the case that $x=1$ or $P(x) = 0$ for those $x$ that satisfy $Q(x) = x$. Besides, I do not see how you are getting that from the equation. why not ? using the second relation right?

But why should $Q(P(x)) = P(x)$?

Kayak wrote: enhanced wrote: Come on guys you all are missing a case,that is since d=0,Q(x)=x^2+cx,now RHS in the condition given in the question is still zero if Q(x)=0 so it has two roots 0,-c so Q(P(-c))=0 and Q(P(0))=0 from here we get two case for the value of b that is b=0 or -c solving these gives more than two sets of P(x) and Q(x) so there are still more roots $Q(x)=0$ obviously doesn't imply $P(Q(x)) = 0$, nor is the converse true. no I said that Q(x)=0 imply that Q(P(x))=0

it doesn't imply , but that's what is asked for ( to find the solutions for I mean)

biomathematics wrote: But why should $Q(P(x)) = P(x)$? if I set $Q(x)=x $ doesn't it imply$ Q(P(x)=P(x)$?

It is given that Q(P(x))=P(x)Q(x) so the LHS is equal to zero for the roots of P and Q

enhanced wrote: Come on guys you all are missing a case,that is since d=0,Q(x)=x^2+cx,now RHS in the condition given in the question is still zero if Q(x)=0 so it has two roots 0,-c so Q(P(-c))=0 and Q(P(0))=0 from here we get two case for the value of b that is b=0 or -c solving these gives more than two sets of P(x) and Q(x) so there are still more roots There are 4 roots, but the the other two are complex not real. The four roots are $-1, \frac{1}{2}, -\frac{1}{4} + \frac{\sqrt{15}i}{4} \text{ and } -\frac{1}{4} - \frac{\sqrt{15}i}{4}$

Wait I will upload my solution

enhanced wrote: Wait I will upload my solution $P(Q(x))$ is a quartic equation it has a maximum of four roots. rd1452002 wrote: The four roots are $-1, \frac{1}{2}, -\frac{1}{4} + \frac{\sqrt{15}i}{4} \text{ and } -\frac{1}{4} - \frac{\sqrt{15}i}{4}$

enhanced wrote: Come on guys you all are missing a case,that is since d=0,Q(x)=x^2+cx,now RHS in the condition given in the question is still zero if Q(x)=0 so it has two roots 0,-c so Q(P(-c))=0 and Q(P(0))=0 from here we get two case for the value of b that is b=0 or -c solving these gives more than two sets of P(x) and Q(x) so there are still more roots What are you saying? Do you write solutions in the exam using the same language? Write proper steps.

AlastorMoody wrote: How's that possible?? The same way INMO 18 P6 was verbatim copy of a Turkey TST problem

I hope so something doesn't happen like that this year too Edit: Lol, and it did happen again, RMO 2018 P6 (i) from FBH-Regional 2012 LOL Edit2: RMO 2018 P6 a copy from INMO 1989

I have a different solution and I am getting an extra solution. Can someone please point out which step I am doing wrong and why. \(P(x)Q(x)=Q(P(x))\) Therfore \(P(a)Q(a)=Q(P(a))\) for '\(a\)' root of \(P(x)\) \(\Rightarrow d = 0 \Rightarrow Q(x) = x^2 + cx\) Now, \(P(0)Q(0)=Q(P(0))\) \(\Rightarrow 0 = b^2 + cb \Rightarrow b = 0 \text{ or } b = -c\) Also, \(P(1)Q(1) = Q(P(1))\) \((\frac{3}{2} + b)(1+c) = (\frac{3}{2} + b)^2 + c(\frac{3}{2} + b)\) \(\Rightarrow b^2 +2b +\frac{3}{4} = 0\) giving two solutions, \(b = -\frac{3}{2} , c = \frac{3}{2}\) \(b = -\frac{1}{2}, c =\frac{1}{2} \) Solving the corresponding equations we get solutions \(\frac{1}{2},-1,-2\).

That's interesting, lemme check

I have a doubt- can't we solve this problem by expanding and comparing coefficients?

Math-wiz wrote: I have a doubt- can't we solve this problem by expanding and comparing coefficients? Yes, we can

Delta0001 wrote: Math-wiz wrote: I have a doubt- can't we solve this problem by expanding and comparing coefficients? Yes, we can Anyone who solved it that way and got full credits? Cuz I got only 5 marks, and I had solved just this one in 2017(I was a dumbo tha time, no change still ). But I guess my solution was incomplete as I did not solve the final equation, but left as it is.

Math-wiz wrote: Delta0001 wrote: Math-wiz wrote: I have a doubt- can't we solve this problem by expanding and comparing coefficients? Yes, we can Anyone who solved it that way and got full credits? Cuz I got only 5 marks, and I had solved just this one in 2017(I was a dumbo tha time, no change still ). But I guess my solution was incomplete as I did not solve the final equation, but left as it is. Yeah, I expanded the whole thing, equated coefficients and got full

Note that, $Q(P(x))=x^4+x^3+(\frac{1}{4} +2b+c)x^2 +(\frac{c}{2} +b)x +(cb+d+b^2) \cdots (1)$. $P(Q(x))=x^4+(2c)x^3+(c^2+d+\frac{1}{2})x^2 +(cd +\frac{c}{2})x +(\frac{d}{2} +b)\cdots (2)$. Equating cofficient of $(1),(2)$ get, $c=\frac{1}{2}$. $\frac{1}{4} +2b+c=c^2+\frac{1}{2} +d$. $or,2b=d$. Again we get , $b^2+\frac{b}{2} =0$ . $or,b=0$ since it is real . So ,$P(x)=x^2 +\frac{x}{2}$ $Q(x)=x^2+\frac{x}{2}$. Clearly $P(Q(x))=(x^2+\frac{x}{2}) (x^2+\frac{x}{2}+1)$. Only two real root exist they are $0,\frac{-1}{2}$.

Sorry I read the question wrong

Note that, $Q(P(x))=x^4+x^3+(\frac{1}{4} +2b+c)x^2 +(\frac{c}{2} +b)x +(cb+d+b^2) \cdots (1)$ $P(x)Q(x)=x^4+(c+\frac{1}{2})x^3+(b+\frac{c}{2}+d) x^2+(\frac{d}{2}+bc)x +bd \cdots (2)$. Equating cofficient of$(1),(2)$ get, From cofficient of $x^3$,$c=\frac{1}{2}$. From cofficient of $x^2$,$b+\frac{1}{4}+d=2b+c+\frac{1}{4}$ which gives $b+\frac{1}{2} =d$. From cofficient of $x^0$ ,get $b^2+\frac{b}{2}+b+0.5=b(b+0.5)$. $or,b= \frac{-1}{2}$ and $d=0$. Get, $P(x) =x^2+\frac{x}{2} -\frac{1}{2}=(x-\frac{1}{2})(x+1)$ $Q(x)=x^2+\frac{x}{2}$. Finally get $P(Q(x))$ has root $1,\frac{-1}{2}$.

Trivial, posting for storage.

smdp wrote: I have a different solution and I am getting an extra solution. Can someone please point out which step I am doing wrong and why. \(P(x)Q(x)=Q(P(x))\) Therfore \(P(a)Q(a)=Q(P(a))\) for '\(a\)' root of \(P(x)\) \(\Rightarrow d = 0 \Rightarrow Q(x) = x^2 + cx\) Now, \(P(0)Q(0)=Q(P(0))\) \(\Rightarrow 0 = b^2 + cb \Rightarrow b = 0 \text{ or } b = -c\) Also, \(P(1)Q(1) = Q(P(1))\) \((\frac{3}{2} + b)(1+c) = (\frac{3}{2} + b)^2 + c(\frac{3}{2} + b)\) \(\Rightarrow b^2 +2b +\frac{3}{4} = 0\) giving two solutions, \(b = -\frac{3}{2} , c = \frac{3}{2}\) \(b = -\frac{1}{2}, c =\frac{1}{2} \) Solving the corresponding equations we get solutions \(\frac{1}{2},-1,-2\). The mistake in this solution is that when b = -3/2, c = -1/2 ; then the required equality is satisfied only for x = 0, 1 ; not for all real x. As a ADVICE, in such questions if you get some value for the variables, don't forget to check whether the required conditions are satisfied or not, since checking takes very less time, but if you have a wrong solution, then it will cost you a lot of marks (and its even more sad when you know how to solve the question but do some silly mistake).

Let $\alpha$ be a root of $P(x)$. Substituting $x=\alpha$ in the given equation, we get that $Q(0)=0\implies d=0$. Now, substituting $x=0$ in the same equation, we get that $b$ is a root of $Q(x)$. Thus, we have $b=-c$, and $Q(x)=x(x-b)$. Now, matching the coefficients of $x^0$, $x$, $x^2$, $x^3$, and $x^4$ in the given equation, we get that $b=-1/2$ is the only possible value of $b$. This gives us $P(Q(x))=\frac{P(x)(2x^2+x+2)}{4}$. Clearly, the only real roots of this polynomial are $x=-1$ and $\frac{1}{2}$. We are done!

anantmudgal09 wrote: I wonder if anyone would've used this idea. Note that $Q(0) \equiv Q(P(x))=P(x) \cdot Q(x) \pmod{P(x)}$ so $Q(0)=0$. Hence $Q(x)=x(x+c)$ while $$x(x+c)P(x)=P(x)(P(x)+c)$$so $P(x)=x^2+cx-c$. Hence $c=\tfrac{1}{2}$ so $Q(x)=x\left(x+\frac{1}{2}\right)$ and $P(x)=\frac{2x^2+x-1}{4}$. Solving for $P(Q(x))=0 \iff t^2+t-2=0$ with $t=x(2x+1)$ we see $x \in \{-1, \tfrac{1}{2}\}$ are the only real roots. I think more people should appreciate how incredibly smart this sol is!