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

## Comments

Sort by:

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

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

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

\(\dfrac{n}{2}((n + 1)(2n + 1) - 3(n + 1) + 1) = \frac{n}{2}(2n^{2} + 3n + 1 - 3n - 3 + 1) = n^{3}.\) – Brian Charlesworth · 1 year, 5 months ago

Log in to reply

– Anupam Nayak · 1 year, 3 months ago

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

– Dev Sharma · 1 year, 5 months ago

Thanks for you solution, sir.Log in to reply

@Brian Charlesworth. Now I will try 4th. – Kushagra Sahni · 1 year, 5 months ago

I solved 3rd very easily and my solution was exactly likeLog in to reply

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. – Sudeep Salgia · 1 year, 5 months ago

Log in to reply

@Surya Prakash @Harsh Shrivastava @Kushagra Sahni @Lakshya Sinha @Nihar Mahajan @Swapnil Das @Adarsh Kumar @Pi Han Goh @Otto Bretscher @Brian Charlesworth @Chew-Seong Cheong – Dev Sharma · 1 year, 5 months ago

Log in to reply

– Nihar Mahajan · 1 year, 5 months ago

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

– Dev Sharma · 1 year, 5 months ago

Why? Reason?Log in to reply

Moderators like Nihar help us to cultivate the Brilliant community, which allows it to flourish :) – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply

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 · 1 year, 5 months ago

Log in to reply

– Dev Sharma · 1 year, 5 months ago

Ok. I will take care of it.Log in to reply