How to find a polynomial with integer coefficients and a root $\sqrt[3]{2} + \sqrt{5}$? The degree of this polynomial can't be more than 7.
121695
Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$  as a root).  I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to  square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals.
25044
I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$.   I managed to find that it is a root of the polynomial $x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17$. This is where I run in to some problems. I don't know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?
111893
Let $f(x)=x^n+5x^{n-1}+3$ where $n\geq1$ is an integer. Prove that $f(x)$ can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree $\geq1$.   If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that $f(x)$ is irreducible over real numbers. But since this condition is given,therefore, we can't exclude the case when $f(x)$ is reducible over real numbers but not with polynomials of integer coefficients. Anyone with idea how to proceed? Thanks in advance!!
10631
Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant.
94686
Suppose that $P(x)$ is a polynomial of degree $n$ such that $$P(k)=\dfrac{k}{k+1}$$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$.    I could not relate this question with this one , so maybe it is not a duplicate  I have absolutely no idea, how the function became a polynomial and how to approach it. Please help!
25125
So there's this set of polynomial expressions with degree n=3:  $$ \left\{ \begin{array}{c} x_1 + x_2 + x_3 = a \\ x_1^2 + x_2^2 + x_3^2 = b \\ x_1^3 + x_2^3 + x_3^3 = c \end{array} \right. $$  How to find atleast one set of x1,x2,x3 values, knowing that all variables and constants (a,b,c) are positive integers?  Thank you.    Update:  Alright, with the help in the comments I was able to transform this set into a single polynomial expression that can be solved in various ways.  Using Newton's identities: $$ \begin{array}{} x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_1x_3) \\ x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 \end{array} $$  We can substitute expressions in the set:  $$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_1x_3) = b \\ (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 = c \end{array} \right. $$  $$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ x_1x_2 + x_2x_3 + x_1x_3 = \frac{a^2 - b}{2} \\ (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 = c \end{array} \right. $$  $$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ x_1x_2 + x_2x_3 + x_1x_3 = \frac{a^2 - b}{2} \\ x_1x_2x_3 = \frac{c + 3a\frac{a^2-b}{2} - a^3}{3} \end{array} \right. $$  To simplify, let's assume (since the right halves of the last set are all constants):  $$ \begin{array}{} p_1 = a \\ p_2 = \frac{a^2 - b}{2} \\ p_3 = \frac{c + 3a\frac{a^2-b}{2} - a^3}{3} \end{array} $$  So we get:  $$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = p_1 \\ x_1x_2 + x_2x_3 + x_1x_3 = p_2 \\ x_1x_2x_3 = p_3 \end{array} \right. $$  Now, using Viete's theorem for general polynomial of degree 3:  $$ P(x) = a_3t^3 + a_2t^2 + a_1t + a_0 $$  With the following properties:  $$ \begin{array}{} t_1 + t_2 + t_3 = -\frac{a_2}{a_3} \\ t_1t_2 + t_2t_3 + t_1t_3 = \frac{a_1}{a_3} \\ t_1t_2t_3 = -\frac{a_0}{a_3} \end{array} $$  Applying to our set:  $$ \left\{ \begin{array}{} p_1 = -\frac{a_2}{a_3} \\ p_2 = \frac{a_1}{a_3} \\ p_3 = -\frac{a_0}{a_3} \end{array} \right. $$  And assuming that $a_3 = 1$, the final polynomial will be the following:  $$ t^3 - p_1t^2 + p_2t - p_3 = 0, $$  where the three roots of $t$  are variables $x_1, x_2$ and $x_3$.  (Using Buchberger's algorithm might be a good way to solve it a well, but I was struggling with it and decided to do it this way for now.)
109759
It's a really basic question,in these days, I've been thinking why a polynomial $p(x)\in F[x]$ ($F$ a field) with degree $n$ can have at most n roots. It seems easy to prove, but I've been trying to prove this since yesterday, maybe I forgot some important details necessary to prove, I don't know. I'm solving some questions about Field Theory and I noticed that almost every question I should use this theorem, I need help.  Thanks
68483
Does there exist a polynomial $f$ with integer coefficients such that for every natural number $n$, $f(0) , f(1) ... f(n) $ are all distinct powers of 2 ?   I have no clue about how should I start thinking about this problem but I felt that it is very interesting.
127
We have a polynomial $x^2 + 3x + 1$. There are 2 zeroes of it,  $\alpha$ and $\beta$  Now, what we need to find out is(What I couldn't) is as follows:  $$ \alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}$$  Any ideas on how to do it?  Thanks a lot! -bone
68065
I saw this problem on a problem set and I have absolutely no idea how to proceed in a feasible way.     Does there exist a polynomial $f(x)$ with real coefficients such that $f(x)^2$ has fewer nonzero coefficients than $f(x)$?   Any help would be appreciated.
122716
$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients.  If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.  I tried by assuming a fifth degree polynomial but got stuck after that.  The question was asked by my friend.
62944
I am looking for an equation     $$x^3+ax^2+bx+c=0, \qquad a, b, c \in \Bbb Z,$$   of degree $3$ that has $3$ different roots.  For an equation of degree $2$ it is easy---for example $x^2-2=0$---but I can't find an example for degree $3$.  if there is a method for finding this kind of Equation I will be grateful.
41115
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$   Suppose $a, b, c$ are roots of $P(x)$.   $$P(x) = k(x - a)(x - b)(x - c)$$  But then I get $(k = 1)$  $$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$  Cant go further...
49801
Can someone please give me a tip how to show that for distinct prime $p,q$ that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ and how to find the minimal polynomial of $\sqrt{p}\sqrt[3]{q}$ over $\mathbb{Q}$ ?
12833
Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated.
70978
Find all polynomials $ f$ with real coefficients such that if $ n$ is a repunit, then so is $ f(n).$  [Note this is a Putnam question, so it is intended to be of easy to middling difficulty as contest problems go.]
37144
I am trying to find the splitting field of $x^6+x^3+1$ over $\mathbb{Q}$.  Finding the roots of the polynomial is easy (substituting $x^3=t$ , finding the two roots of the polynomial in $t$ and then taking a 3-rd root from each one). The roots can be seen  [if there is a more elegant way of finding the roots it will be nice to hear]  Is is true the that the splitting field is $\mathbb{Q}((-1)^\frac{1}{9})$ ? I think so from the way the roots look, but I am unsure.  Also, I am having trouble finding the minimal polynomial of $(-1)^\frac{1}{9}$, it seems that it would be a polynomial of degree 9, but of course the degree can't be more than 6...can someone please help with this ?