I posted  a while back, and I think I may not have made myself clear or shown what I got so far. So let me post this question again with more information and clarification.   I have a sequence of polynomials $Q_k(x, y)$, $k\geq 1$ defined recursively as follows:   $Q_1=x$. There is a sequence of polynomials $p_j(y)$ of degree $j$ such that $Q_{2m}$ is of the form \begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray} and $Q_{2m+1}$ \begin{eqnarray} \frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x \end{eqnarray} $Q_k(k+1-2i, k+1)$ as a polynomial in $i$ has roots $1, 2, \cdots, k$.    Find a general formula of $Q_k(x, y)$. The following are the first 4 polynomials in the sequence: \begin{align*} Q_1&amp;=x\\ Q_2&amp;=\frac{x^2}{2}-\frac{y}{6}\\ Q_3&amp;=\frac{x^3-xy}{6}\\ Q_4&amp;=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360} \end{align*}  In fact, it suffices to find the sequence of polynomials $p_j(y)$. Here's what I've got so far. By condition (2) and (3), we have \begin{eqnarray} \sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m) \end{eqnarray} If we let $m=j+k$, differentiate both sides with respect to $i$ $2k$ times and put $\displaystyle i=\frac{2j+2k+1}{2}$, we have \begin{eqnarray} p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k)) \end{eqnarray} Letting $k=0, 1, \cdots, j$, we get the $j+1$ values taken by $p_j$ at $2j+1, 2j+3, \cdots, 4j+1$. $p_j$ then can be computed using Lagrangian interpolation.   It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula?  Edit: The first 5 members in the polynomial sequence $p_j(y)$ are the following: \begin{align*} p_0(y)&amp;=1\\ p_1(y)&amp;=-\frac{y}{6}\\ p_2(y)&amp;=\frac{y(5y+2)}{360}\\ p_3(y)&amp;=-\frac{y(35y^2+42y+16)}{45360}\\ p_4(y)&amp;=\frac{y(5y+4)(35y^2+56y+36)}{5443200} \end{align*}
123748
I have a sequence of polynomials $Q_k(x, y)$, $k\geq 1$ defined recursively as follows:   $Q_1=x$. There is a sequence of polynomials $p_j(y)$ of degree $j$ such that $Q_{2m}$ is of the form \begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray} and $Q_{2m+1}$ \begin{eqnarray} \frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x \end{eqnarray} $Q_k(k+1-2i, k+1)$ as a polynomial in $i$ has roots $1, 2, \cdots, k$.    Find a general formula of $Q_k(x, y)$. The following are the first 4 polynomials in the sequence: \begin{align*} Q_1&amp;=x\\ Q_2&amp;=\frac{x^2}{2}-\frac{y}{6}\\ Q_3&amp;=\frac{x^3-xy}{6}\\ Q_4&amp;=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360} \end{align*}  In fact, it suffices to find the sequence of polynomials $p_j(y)$. Here's what I've got so far. By condition (2) and (3), we have \begin{eqnarray} \sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m) \end{eqnarray} If we let $m=j+k$, differentiate both sides with respect to $i$ $2k$ times and put $\displaystyle i=\frac{2j+2k+1}{2}$, we have \begin{eqnarray} p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k)) \end{eqnarray} Letting $k=0, 1, \cdots, j$, we get the $j+1$ values taken by $p_j$ at $2j+1, 2j+3, \cdots, 4j+1$. $p_j$ then can be computed using Lagrangian interpolation.   It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula?  Edit: The first 6 members in the polynomial sequence $p_j(y)$ are the following: \begin{align*} p_0(y)&amp;=1\\ p_1(y)&amp;=-\frac{y}{6}\\ p_2(y)&amp;=\frac{y(5y+2)}{360}\\ p_3(y)&amp;=-\frac{y(35y^2+42y+16)}{45360}\\ p_4(y)&amp;=\frac{y(5y+4)(35y^2+56y+36)}{5443200}\\ p_5(y)&amp;=-\frac{y(385y^4+1540y^3+2684y^2+2288y+768)}{359251200} \end{align*}  It is noteworthy that the denominators happen to be the first 6 numbers in , as pointed out by Solomonoff's Secret.
5078
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$  I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.  Link:
64934
Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $&lt;n$. Once again, the task is to show  $$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$  For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$.  I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?
40763
I am stuck with the following problem. Every polynomial of degree $d$ can be expressed as $$ p(x) = p_d \binom{x}{d}+ p_{d-1}\binom{x}{d-1} + \cdots + p_0 \binom{x}{0} $$     What is the representation of    $$ \binom{x}{i} \binom{x}{j}  $$   in such a basis?   I have no idea how to solve this...
109935
Find all polynomials $p$ with complex coefficients such that                                    $$ p(x^2)=p(x)p(x+1).$$   The goal is to find a general formula for polynomials that satisfy the above equation. I believe the answer is $\left(x^2-x\right)^k$ with $k\in\mathbb{N}$, but I have no idea how to get this solution.
105456
Let $P(x)$ be a degree $n$ polynomial with distinct roots $r_1, r_2, \cdots r_n$. Prove that   $$\displaystyle\sum_{i=1}^{n}\frac{P''(r_i)}{P'(r_i)}=0$$  My proof: We can equivalently rewrite the polynomial as $(x-r_1)(x-r_2)\cdots(x-r_n)$ and if we plug in the roots in the polynomial, we get that all of them are equal to $0$ so their first derivatives are equal to $1$ and then their second derivatives are equal to $0$ which means that all quotients are equal to $0$ and there for the sum is as well, $Q.E.D$  Don't show the solution please, but if you have a hint, feel free. Thank you.
83964
Let $f \in \mathbb{Q}$ be a polynomial of degree $n&gt;0$. Let $p_1,p_2,p_3..p_{n+1}$ be distinct prime numbers. Show that there exists a non-zero polynomial $g \in \mathbb{Q}[X]$ such that $fg=\sum_{i=1}^{n+1}c_iX^{p_i} $ with $c_i \in \mathbb{Q}$.  If I want to find a polynomial then it must be of degree $p_{n+1}-n$. So I assumed a polynomial $g=\sum_{i=1}^{p_{n+1}-n}b_iX^{i}$.  Multiplying $f$ &amp; $g$ and equating the non-prime power coefficents to $0$ where our variables are $b_i's$ (they are $p_{n+1}-n+1$ in number) and we have $p_{n+1}-n+1$ equations. One solution is all $b_i=0$ and in this case it is zero polynomial.   How do I proceed further? Is there another way to prove it?
66019
Find polynomials $A(x)$ and $B(x)$ such that $A(x)P(x) + B(x)Q(x) = x + 1$ for all $x$ where $P(x) = x^4 - 1$ and $Q(x) = x^3 + x^2$. I'm stumped on this question. I know that I'm supposed to apply the extended version of Euclid's algorithm for polynomials but I'm unsure of how to do that. I thought about trying to create some kind of linear system but guessing arbitrary coefficients but that wouldn't work as $A$ and $B$ don't have fixed degrees.
88370
Prove that if $p$ is a polynomial of degree at most $n-1$, then   $$\sum_{k=0}^n \binom{n}{k}(-1)^k p(k) = 0.$$  This was probably asked before but I couldn't find it anywhere. Can anyone please help me out here? Looks like the proof is extensive.
53386
I have to find $P(0)$ from the polynomial with minimum degree given that  $$(x-1)^3|(P(x)+1)$$  $$(x+1)^3|(P(x)-1)$$  Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively:  $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$  Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$  and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.
27111
Suppose $P(x)$ is a polynomial of degree $n \geq 1$ such that $\int_{0}^{1}x^{k}P(x)\,dx = 0$ for $k = 1, 2, \ldots, n$. Show that $$\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$$   If we assume that $P(x) = a_{0}x^{n} + \cdots + a_{n - 1}x + a_{n}$ then we can easily see that $$\int_{0}^{1}\{P(x)\}^{2}\,dx = a_{n}\int_{0}^{1}P(x)\,dx$$ and therefore to solve the given problem we need to show that $$\int_{0}^{1}P(x)\,dx = \frac{a_{n}}{(n + 1)^{2}}$$ Direct integration of the polynomial gives the expression $$\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \cdots + \frac{a_{n - 1}}{2} + a_{n}$$ and simplifying this to $a_{n}/(n + 1)^{2}$ does not seem possible. I think there is some nice "integration by parts" trick which will give away the solution, but I am not able to think of it.
371
I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:   $f(x)-1$ is divisible by $(x-1)^3$ $f(x)$ is divisible by $x^3$   To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$. This is divisible by $x^3$, so $d, e, f $  must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$  This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.  From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
14055
I woke up and suddenly wondered about the following: A polynomial of degree $n$ $a_n x^n + a_{n-1} x^{n-1} + \cdot\cdot\cdot + a_{n-n} x^{n-n}$ can be written as $a_n x^n + g(x)$ where $g(x)$ is a polynomial of degree $n-1$. Then I was thinking that for every $g(x)$ has a unique set of roots with respect to their values and multiplicities - is this true?  So for instance consider a polynomial of degree $2$: $x^2 + a_1 x + a_0$ which can be written as $x^2 + g(x)$ where $h(x) = a_1 x + a_0$. Then every unique combination of $a_1$ and $a_0$ in $g(x)$ will give a unique set of roots in terms of values and multiplicities.   An example of this is $x^2 + 5x + 4$ (with $a_2 = 1$) - this polynomial has the roots $-4$ and $-1$, both with a multiplicity of $1$. Thus, for this $a_2$, only $g(x) = 5x + 4$ will give this set of roots. Only for a different $a_n$ a different $g(x)$ may yield this set of roots. I mean after all, any polynomial of degree $2$ can be written as $a_2(x - r_1)(x - r_2)$.  This is all something that I think could be true. So can any of you smart guys out there tell me if I am right and if so, might explain why this holds - that the set of roots with respect to their values and multiplicities are unique for each $g(x)$?
88170
Find all polynomials $P:\mathbb{C}\rightarrow\mathbb{C}$ such that $$P(x^2)=P(x)^2 .$$   Here is what I tried:  First, it is easy to see the constant solutions, namely $P\equiv 0,P\equiv 1$.  Let $r$ be a root of $P$ (i.e. $P(r)=0$). It follows that all terms in the infinite sequence $r,r^2,r^4,\dots,r^{2n},\dots$ are roots of $P$. To avoid a polynomial with infinite roots, we have to have the sequence be periodic. So it turns out that either $r=0$, or the roots of $P$ are roots of unity of degree $2n$ for some $n\in\mathbb{N}$.  In the first case, we get $P(x)=xQ(x)$. Plugging in, we get $$x^2Q(x^2)=x^2Q(x)^2\implies Q(x^2)=Q(x)^2 ,$$ i.e. that $Q$ satisfies the same condition as $P$.  In the second case, we have $P(x)=(x^{2n}-1)Q(x)$. Plugging in, we get $(x^{2n}-1)Q(x^2)=(x^{2n}-2x^n+1)Q(x)^2$. I was not sure how to proceed from here.
43243
Consider the following family of polynomials $$P_n(X)=\sum_{i=0}^n(n+1-i)X^i,\,n\ge 1$$ Let’s write down the first few $$ \begin{align} &amp;P_1(X)=X+2\\ &amp;P_2(X)=X^2+2X+3\\ &amp;P_3(X)=X^3+2X^2+3X+4\\ &amp;P_4(X)=X^4+2X^3+3X^2+4X+5 \end{align} $$ My claim is that this family is a family of irreducible polynomials in $\Bbb{Z}[X]$. I proved it for $n \le 5$ by the Eisenstein criterion after the change of variable $X=t+1$ For $n=6$ after the change of variable the polynomial writes as follows $$Q_6(t)=P_6(X-1)=t^6+8t^5+28t^4+56t^3+70t^2+56t+28$$ and the Eisenstein criterion (only workable $p=2$) doesn’t work anymore. By reducing $\bmod 7$ we prove the claim for $n=6$. I tested with Mathematica and irreducibility is checked for $n\le 150$. I noticed $$P_n(X)=XP_{n-1}(X)+n+1$$ But I am struggling to find a generic proof. I have given up on counterexample. Thanks for your help.
37333
I would very much like to have a complete list of the types of polynomial functions. I know that theres:  Quadratic :                      (AX^2 + BX + C) Cubic     :               (AX^3 + BX^2 + CX + D) Quartic   :        (AX^4 + BX^3 + CX^2 + DX + E) Quintic   : (AX^5 + BX^4 + CX^3 + DX^2 + EX + F)   What are the names of polynomial functions to the further powers? Sexstic? Septic? Octic? I'm not sure, so if someone could enlighten me, then that would be great. Please provide a list that goes at least to the seventh power, but it would be nice if you could go further. Thanks.
53764
Find all real polynomials with $P(0)=0$ and $P(x^2+1)=(P(x))^2+1$  I' like a solution for this problem  and proof verification:  My proof:      Let $a_0=0$ and $a_{n+1}=a_n^2+1$, clearly $P(a_0)=a_0$ and if $P(a_n)=a_n$ then $P(a_{n+1})=P(a_n^2+1)=(P(a_n))^2+1=a_n^2+1=a_{n+1}$. So $P(x)=x$ for infinitely many values of $x$ (because $a_n$ is strictly increasing), hence $P$ is the identity polynomial, and it clearly works.