Olympiad Corner

3) Let P(x)P(x) be second degree polynomial such that : P(1)=1P(1)=1 P(2)=7P(2)=7 P(3)=19P(3)=19 show that P(1)+P(2)+P(3)+......+P(n)=n3P(1) + P(2) + P(3) + ...... + P(n) = n^3

4) Find all pairs of polynomials P(x)P(x) and Q(x)Q(x) with real coefficients such that : P(x2+1)=Q(x)2+2xP(x^2 + 1) = Q(x)^2 + 2x and Q(x2+1)=P(x)2Q(x^2 + 1) = P(x)^2

Different approaches will be appreciated

Note by Dev Sharma
3 years, 11 months ago

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3) With P(x)=ax2+bx+cP(x) = ax^{2} + bx + c we have the system

(i)a+b+c=1,(ii)4a+2b+c=7,(iii)9a+3b+c=19.(i) a + b + c = 1, (ii) 4a + 2b + c = 7, (iii) 9a + 3b + c = 19.

Now (ii) - (i) yields 3a+b=6,3a + b = 6, and (iii) - (i) yields 8a+2b=18.8a + 2b = 18.

Thus (8a+2b)2(3a+b)=1826=62a=6a=3,b=3,c=1.(8a + 2b) - 2*(3a + b) = 18 - 2*6 = 6 \Longrightarrow 2a = 6 \Longrightarrow a = 3, b = -3, c = 1.

So k=1nP(k)=k=1n(3k23k+1)=\displaystyle\sum_{k=1}^{n} P(k) = \sum_{k=1}^{n} (3k^{2} - 3k + 1) =

3k=1nk23k=1nk+k=1n1=\displaystyle 3*\sum_{k=1}^{n} k^{2} - 3*\sum_{k=1}^{n} k + \sum_{k=1}^{n}1 =

3n(n+1)(2n+1)63n(n+1)2+n=3*\dfrac{n(n + 1)(2n + 1)}{6} - 3*\dfrac{n(n + 1)}{2} + n =

n2((n+1)(2n+1)3(n+1)+1)=n2(2n2+3n+13n3+1)=n3.\dfrac{n}{2}((n + 1)(2n + 1) - 3(n + 1) + 1) = \frac{n}{2}(2n^{2} + 3n + 1 - 3n - 3 + 1) = n^{3}.

Brian Charlesworth - 3 years, 11 months ago

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Thanks for you solution, sir.

Dev Sharma - 3 years, 11 months ago

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I solved 3rd very easily and my solution was exactly like @Brian Charlesworth. Now I will try 4th.

Kushagra Sahni - 3 years, 11 months ago

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I used method of difference to find the polynomial and then did the second part the same as you.

Anupam Nayak - 3 years, 9 months ago

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The polynomials P(x) P(x) and Q(x)Q(x) have the same degree and are monic. This is easily inferred by assuming degrees to be mm and nn and then equating the degree of the polynomials thus formed in the second equation. Now it is important to realize that all the terms in the polynomial P(x2+1) P(x^2 +1) will be ones with an even power of xx. Thus so must Q(x)2+2x Q(x)^2 + 2x . Thus Q(x)2 Q(x)^2 must have a 2x - 2x in its expansion. Also it must be the only term with a odd power of xx. Such a term in the expansion indicates that the coefficient of xx and the constant term in Q(x) Q(x) are non-zero. I'll give a rough sketch as to why should these be the only nonzero coefficients. Assume Q(x)=i=0naixi\displaystyle Q(x) = \sum_{i=0}^n a_i x^i such that a0,a1,an0a_0 , a_1 ,a_n \neq 0 . Consider the coefficient of x2n1 x^{2n-1} in the expansion of Q(x)2Q(x)^2. This will have to be zero for n>1 n >1. It implies an1=0a_{n-1} =0 . With a similar argument for x2n3 x^{2n-3} , we can conclude that an3=0 a_{n-3} = 0. This can be continued until a0a_0 or a1a_1 is reached depending upon nn, which would force either ana_n to be 00 or a0/a1 a_0 / a_1 to be 00. Since the latter condition is more important for the given condition to hold true, all the conditions, when imposed together will end up with n=1n=1, i.e., Q(x)Q(x) is linear. After this, it is easy to realize that P(x)=x P(x) = x and Q(x)=x1 Q(x) = x - 1. I think these are the only possible polynomials.

Note: I haven't given the complete proof. It is rough sketch which can be made more rigorous and complete with more logical flow of arguments all of which I have mentioned. If you think I have missed a step, let me know I'll make myself more clear.

Sudeep Salgia - 3 years, 11 months ago

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This is a warning: Limit your tags to at most 5 people.

Nihar Mahajan - 3 years, 11 months ago

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Why? Reason?

Dev Sharma - 3 years, 11 months ago

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@Dev Sharma That is a community standard that we adhere to. Often, I would reply to such mass tagging with a comment like

Please refrain from tagging too many people. You should choose at most 5 people who are the most likely to be interested and will respond to your note.

Moderators like Nihar help us to cultivate the Brilliant community, which allows it to flourish :)

Calvin Lin Staff - 3 years, 10 months ago

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@Dev Sharma Hey friend , people have other work to do too. If you want those people to see your note , as time passes , eventually this note will be seen by them and they would respond accordingly. If many people on brilliant start mass tagging of smart people like Cheong sir or Otto Bretscher or Pi Han or Brian sir or others , it would be trouble for them to respond all the notes individually.Almost every person wants his note to be seen by all these members which is hectic. I hope you get the point.

So please limit your tags to atmost 5 people. This is not my instruction , this is the common instruction of brilliant staff. Thanks :)

Nihar Mahajan - 3 years, 11 months ago

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@Nihar Mahajan Ok. I will take care of it.

Dev Sharma - 3 years, 11 months ago

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