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

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

Different approaches will be appreciated

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## Comments

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TopNewest3) With \(P(x) = ax^{2} + bx + c\) we have the system

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

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

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

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

\(\displaystyle 3*\sum_{k=1}^{n} k^{2} - 3*\sum_{k=1}^{n} k + \sum_{k=1}^{n}1 =\)

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

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

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Thanks for you solution, sir.

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

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

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

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@Surya Prakash @Harsh Shrivastava @Kushagra Sahni @Lakshya Sinha @Nihar Mahajan @Swapnil Das @Adarsh Kumar @Pi Han Goh @Otto Bretscher @Brian Charlesworth @Chew-Seong Cheong

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

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Why? Reason?

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

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

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