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

Note by Dev Sharma
5 years, 8 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

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

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

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

- 5 years, 8 months ago

Thanks for you solution, sir.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 6 months ago

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.

- 5 years, 8 months ago

- 5 years, 8 months ago

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

- 5 years, 8 months ago

Why? Reason?

- 5 years, 8 months ago

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 :)

Staff - 5 years, 8 months ago

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 :)

- 5 years, 8 months ago

Ok. I will take care of it.

- 5 years, 8 months ago