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15

Here's an example. Let $\zeta=e^{2\pi i/7}$ and let $p(x)=(\zeta-x)(\zeta^2-x)(\zeta^4-x)$. Since $\zeta$ is a root of $p$ but $\bar{\zeta}$ is not, $p$ does not have real coefficients. Now notice that $(\zeta^4)^2=\zeta$, so the squaring map permutes the roots of $p(x)$. We can use this to compute \begin{align*} p(x^2)&=(\zeta-x^2)(\zeta^2-x^2)(\...


9

Another way of saying that $f(x)|f(x+f(x))$ is that $f(x+f(x))$ belongs to the ideal $(f(x))$ generated by $f(x)$ in $K[x]$. At the same time this is equivalent to $\pi(f(x+f(x))=\overline{f(x+f(x))}=0$, where $\pi$ is the natural quotient homomorphism $$\pi:K[x]\to K[x]/(f(x)).$$ Now you're done cause $$\overline{f(x+f(x))}=f(\overline{x}+\overline{f(x)})=f(...


7

Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get $$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} $$ ...


6

Note that $Q(x) = (x-1)^2$, which means that $(x-1) \mid P(x)$ and $(x-1) \mid P'(x)$. In other words $x=1$ is a zero of both $P(x)$ and $P'(x)$. So after all you're left to solve the system of linear equation: $$\begin{cases} a-b+1 = 0 \\ 2014a - 2015b = 0 \end{cases}$$ It's easy to conclude that $a=-2015$ and $b=-2014$


5

$2x^4+x^3-11x^2+x+2=0$ Note that the coefficients: $2,1,-11,1,2$ are symmetrical. $2(x^4+1)+(x^3+x)-11x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$ $2(x+1)^4-7(x+1)^2x-9x^2=0$ $2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$


4

Because we are in characteristic $2$ we have $$\phi(X)^2=\left(\sum_{k=0}^{n-1}X^{2^k}\right)^2=\sum_{k=0}^{n-1}(X^{2^k})^2=\sum_{k=1}^nX^{2^k}=\phi(X)+(X^{2^n}-X).$$ As we have $x^{2^n}-x=0$ for all $x\in\Bbb{F}_{2^n}$ it follows that $\phi(x)^2=\phi(x)$ holds for all $x\in\Bbb{F}_{2^n}$, and hence $\phi(x)\in\Bbb{F}_2\subset\Bbb{F}_{2^n}$. For the second ...


4

The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, ...


4

We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of $$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$ $$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$ $$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$ simplify giving the coefficients of $x^...


4

Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$. If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then its derivative is $2015ax^{2014}-2014bx^{...


4

$$2x^4+x^3-11x^2+x+2 = (x-2)(2x^3+5x^2-x-1)=(x-2)(2x-1)(x^2+3x+1)$$ Now you can solve it easily EDIT: You can use the rational root theorem to get the possible rational roots


3

If you want an expression that holds for all $n \geq 2$ then it might be a tad ugly in Cartesian, because: $$\cos(n\theta) = \sum_{\text{even }k} (-1)^{k/2}{n \choose k}\cos^{n-k} \theta \sin^k \theta \\ = (x^2+y^2)^{-n/2}\sum_{\text{even }k} (-1)^{k/2}{n \choose k} x^{n-k}y^k.$$ But this will get you there.


3

This is false. For example, $X^2+2$ and $X^2+6$ are both equal and irreducible mod $4$. However, since $\mathbb Q_2$ does not contain a square root of $3$, their roots give different extensions of $\mathbb Q_2$.


3

We can start by proving that for all $k\ge1$ $$ f_k(x)=k!\cdot(x-x_0)\cdots(x-x_k) $$ where $$ 0=x_0=x_1<x_2<\cdots<x_k<1. $$ Let's assume this holds for $f_k$ and prove it for $f_{k+1}$. First, we may note that the derivative $$ f_k'(x)=(k+1)!\cdot(x-y_1)\cdots(x-y_k) $$ where $$ 0=x_0=y_1=x_1<y_2<\cdots<y_k<x_k<1. $$ The $y_1=0$ ...


3

In A Note on Trigonometric Algebraic Numbers by D. H. Lehmer (and also here with a different notation) we find that $$ z^{-d}\Phi_n(z) = \psi_n(z+z^{-1}) $$ where $\Phi_n$ is the $n$-th cyclotomic polynomial and $d=\frac{\phi(n)}{2}$ is the degree of $\psi_n$, which is half the degree of $\Phi_n$. Lehmer proves that $\psi_n$ is irreducible. The roots of $\...


3

By the rational root theorem, there are $6$ possibilities to check: $\pm\frac12,\pm1,$ or $\pm2$. The easiest values to test are $\pm1$ and neither works. The next easiest value to test is $2$, which is a solution. In addition, using Kenny Lau's observation from his comment, the equation can be re-written as $$2x^2+x-11+\frac1x+\frac2{x^2}=0$$ Since $x=...


3

It is false. Consider the matrix.$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ It is not the square of a complex matrix, so $p(X)=X^2$ is not surjective.


3

Try to "guess" some rational root $\;\cfrac rs\;$ , which by the Rational Root Theorem must fulfill $\;r\,\mid\,-2\;,\;\;s\,\mid\,1\;$ , and indeed $\;2\;$ is a root, so divide by $\;x-2\;$ : $$x^3-3x-2=(x-2)(x^2+2x+1)=(x-2)(x+1)^2$$ and you have one simple root and one double one. If there is no rational root then the task is much, but really much harder ...


2

Hint $\ {\rm mod}\ x^2-1\!:\,\ x^2\equiv 1\,\Rightarrow\,p(x)\equiv x-1 + x-1 + x-1 \equiv 3(x-1)$


2

For the first limit: Recognize that $$\lim_{n\rightarrow \infty}\frac{7n^5-2}{(n+4)^5n}$$ Can really be replaced by the higher order terms, since $n^m$ is way smaller than $n^{m+1}$ when $n$ is really big. So you can see that this limit should be zero, since when you expand the denominator, you will get something with an $n^6$ term. $$\lim_{n\rightarrow \...


2

Being a finite sum, this one converges for every $x \in \Bbb C$, so its radius of convergence is $\infty$.


2

Your argument of a) is incomplete. You proved that $P(X)$ has no root in $\mathbb Q$, and to draw the conclusion it's necessary to add the proof that $P(X)$ has no factor of degree $2$, which is pretty similar to your original argument. In fact, if you know that a monic polynomial $f(X)\in\mathbb Z[X]$ is irreducible iff it's irreducible in $\mathbb Q[X]$, ...


2

We have that: $$A^n+A^{n-1}+...+1=\frac{A^{n+1}-1}{A-1} \equiv (A^{n+1}-1)(A-1)^{-1} \pmod{10^9+7}$$ We know that we can use the Extended Euclidean algorithm to find $(A-1)^{-1}$ because $10^9+7$ is prime, so the only way this wouldn't work is if $A-1$ is a multiple of $10^9+7$, which almost never happens. If we have $1 < A \leq 10^9$, then it definitely ...


2

Every Euclidean domain is a principal ideal domain, but the ideal $(x,y)$ in $F[x,y]$ is not principal.


2

Every Euclidean domain is a principal ideal domain. Yet $F[x,y]$ is not a principal ideal domain, and thus cannot be Euclidean.


2

No. As the other answers point out, it's not a P.I.D. However, it's a U.F.D.; i.e., any polynomial in two or more indeterminates has a factorisation into a product of irreducible polynomials, which is unique up to a unit and the order of the factors. Furthermore, if we fix a monomial order on $F[x_1,\dots,x_n]$ (i.e. a total order on monomials compatible ...


2

For a); not having a root does not imply irreducibility. For example, $X^4+2X^2+1$ has no real roots, but it is certainly reducible. In stead consider using Eisenstein's criterion. For b); it is not true that $\alpha=q+(P(X))$ for some $q\in\Bbb{Q}$. In stead, $\alpha=Q(X)+(P(X))$ for some $Q(X)\in\Bbb{Q}[X]$. If you have seen enough theory, you might say ...


2

For the polynomial $p(x) = x^n + a_1x^{n-1} + \dots + a_n$ we have that the companion matrix $C(p)$ of $p$ given by $$ C(p) = \begin{bmatrix}0&0&\cdots &0 & -a_n\\ 1&0&\cdots & 0 & -a_{n-1}\\0&1&\cdots &0 & -a_{n-2}\\\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&-a_1\end{...


2

By the translation $x\to x-\frac13$, you cancel the square term. $$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$ Then multiplying by $27$ and replacing $x\to3x$, you get $$x^3-57x-133=0.$$ Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\theta)$ appear), $$152\sqrt{19}\cos^...


2

Have you tried any examples? Simplest example to try: $P(z)=z$.



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