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30

Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. $$\require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac ...


20

This is to make nice rules such as $$ \text{deg }(PQ) = \text{deg }P + \text{deg }Q\\ \text{deg }(P+Q) \le \max(\text{deg }P , \text{deg }Q) $$ So the only value that makes it possible is $$\text{deg }0= -\infty$$


15

Define $g(x)=\frac{a_n}{n+1}x^{n+1}+\dotsb+\frac{a_1}2x^2+a_0x$ then $g(0)=g(1)=0$,hence by Rolle's theorem there is some $c$ in the interval $(0,1)$ such that $g'(c)=0$ as we want.


12

Since the polynomial $x^3-x-1$ is irreducible over$\def\Q{\Bbb Q}~\Q$ by the rational root test, one approach would be to identify the element $\frac{1+a}{1-a}$ where $a$ is the image of $x$ in the field $K=\Q[x]/(x^3-x-1)$, and to compute its minimum polynomial over$~\Q$; since the Galois group of the splitting field of $x^3-x-1$ permutes its roots ...


12

Without loss of generality, we may assume that $p$ is monic. Since $p$ has no real roots, $n=2m$ for some $m\ge 1$, and there exist quadratic monic polynomials $q_1,\dots,q_m$ with no real roots such that $p=\prod_{k=1}^m q_k$. Therefore, by Cauchy-Schwarz inequality, $$ \left(\frac{p'}{p}\right)^2=\left(\sum_{k=1}^m \frac{q_k'}{q_k}\right)^2\le m\cdot ...


9

The polynomial has no real roots, because it is equal to $(x^2-2)^2+12$. The remaining possibility is thus that it is a product of two quadratic factors. By Gauss' Lemma these need to have integer coefficients, so we are looking for a possibile factorization like $$ p(x)=x^4-4x^2+16=(x^2+ax+b)(x^2+cx+d) $$ with some integers $a,b,c,d$. Modulo $3$ we have the ...


9

The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square: \begin{align} x^4-4x^2+16 &=x^4+8x^2+16-12x^2\\ &=(x^2+4)^2-(\sqrt{12}\,x)^2\\ ...


8

Below is an explicit proof. Note that $x^4-4x^2+16 = (x^2-2)^2 + 12$, which clearly has no real root. Hence, the only possible way to reduce $x^4-4x^2+16$ over $\mathbb{Q}$ is $(x^2+ax+b)(x^2+cx+d)$. However, the roots of $x^4-4x^2+16$ are $x = \pm \sqrt{2 \pm i\sqrt{12}}$, which are all complex numbers. Since complex roots occur in conjugate pairs, $\sqrt{2 ...


8

Key Idea $\ $ Completing the square leads to a difference of squares $$\begin{eqnarray} \color{#0a0}{x^4+1} -x^2 &\,=\,&\!\!\! \overbrace{\color{#0a0}{(x^2\!+1)^2}}^{\rm\!\!\! complete\ the\ \color{#0a0}{square}\!\!\!}\!\!\!\!\!-\!(\color{#c00}{\sqrt 3 x})^2\ \ \text{so, factoring this} \it\text{ difference of squares}\\ &\,=\,& ...


7

Assigning a degree to the zero polynomial will cause trouble with important and useful theorems that relate the degree of a polynomial to its roots: If $F$ is a field (examples of fields are $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$, $\mathbb{Z}/p\mathbb{Z}$), a polynomial $P$ with coefficients in $F$ (the set/ring of these polynomials is usually denoted by ...


5

Start with $x=\sqrt 2+\sqrt 3$ and square both sides: $$x^2=2+2\cdot\sqrt 6+3 $$ Now isolate the square root and square again: $$(x^2-5)^2=(2\cdot\sqrt 6)^2=24 $$ Expand. (Why is the resulting polynomial irreducible?)


5

The key is that $\mathbb{Z}_4$ (the ring of residue classes modulo $4$) has nilpotent elements. Suppose $A$ is a commutative ring and that $a$ is a nilpotent element; then $1-a$ is a unit in $A$, because $$ 1-a^n=(1-a)(1+a+a^2+\dots+a^{n-1}) $$ for all $n>1$, and we can take $n$ such that $a^n=0$. Since $[2]$ (the residue class of $[2]$ modulo $4$) is ...


5

Actually you have: $$x^4-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt3 x)^2 $$ and use the identity $a^2-b^2=(a-b)(a+b)$


5

The first to try would be to look for (rational) roots, but that is fruitless here (you need only test divisors of the constant term, but $\pm1$ is not a root). Next you might try to factor as $(x^2+ax+b)(x^2+a'x+b')$ with integer coefficients, where once again you could conclude that $bb'=1$, so $b=b'=\pm1$. However, as the solutionm tells as, this won't ...


5

You've seen that $f(x)$ has no roots, so you want to exclude factorizations of the form $$f(x) = (x^2 + ax + b)(x^2 + cx + d)$$ Since $f(x) = f(-x)$, the above implies $$f(x) = (x^2 - ax + b)(x^2 - cx + d)$$ Here $a,b,c$, and $d$ are integers by Gauss's Lemma. So a given root $r$ of $x^2 - ax + b$ is a root of $x^2 + ax + b$ or $x^2 + cx + d$. If $r$ is a ...


5

Let $$f(x)=a_0+a_1x+\ldots +a_nx^n$$ Let $p,q$ be two distinct odd primes. Then because $\gcd(p,q)=1$ we find $u_k,v_k$ such that $a_k=u_kp+v_kq$ and thus letting $$ g(x)=(u_0+rq)p+u_1px+\ldots +u_npx^n+x^{n+1}\\ h(x)=(v_0-rp)q+v_1qx+\ldots +v_nqx^n-x^{n+1}\\$$ (for some integer $r$) we have clearly that $f(x)=g(x)+h(x)$. Also, $g$ and $h$ do almost satisfy ...


5

1. Notation and the Main Statement For each $\mathrm{z} \in \Bbb{C}^d$, we define $$ p(t) = p(\mathrm{z}, t) = \prod_{j=1}^d (t - z_j), \qquad I = I(\mathrm{z}) = \int_{-\infty}^{\infty} \frac{p_{\mathrm{z}}'(t)^2}{p_{\mathrm{z}}(t)^2 + p_{\mathrm{z}}'(t)^2} \, \mathrm{d}t \tag{1} $$ whenever the denominator of the integrand does not vanish. Also by ...


4

The coefficients of $P_n(x)$ are called the Eulerian numbers and there is a corresponding Euler's triangle. They satisfy $$ \genfrac{\langle}{\rangle}{0}{}{n}{k} = (k+1)\genfrac{\langle}{\rangle}{0}{}{n-1}{k}+(n-k)\genfrac{\langle}{\rangle}{0}{}{n-1}{k-1}, $$ for integer $n>0$ as well as $$ ...


4

You can get it by$$x^4-x^2+1=x^4+2x^2-2x^2-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt 3x)^2.$$


4

No, it's not correct. $x^5 = x$ in $\mathbb Z_5$, so $x^j = x^k$ if $j \equiv k \mod 4$ (not mod $5$) and $j,k > 0$.


4

The extreme point is a maximum or minimum, in the quadratic function case it will be global maximum of minimum. If you know calculus, this is easy to understand: at the extreme point, the first derivative at that point is $0$. Differentiate $ax^2+bx+c$ you get $2ax+b=0$, thus $x = -\frac{b}{2a}$ If you don't know calculus, then this is the way you may ...


4

Hint Considering $$3.8r - 0.057r^2 + 0.00038r^3 + 0.00000095r^4$$ start defining $r=100 t$ to get rid of all the decimals. So, the expression becomes $$95 t^4+380 t^3-570 t^2+380 t=95(t^4+4t^3-6t^2+4t)$$ Now, setting the above expression equal to $95$, the problem reduces to $$t^4+4t^3-6t^2+4t-1=0$$ which is a quartic that you can solve with radicals. It ...


4

Hint: the term $\frac{1}{n+1}, \frac 1n,\cdots$ is a hint that you can do integration. Integrate the polynomails on $[0,1]$ and see what happens.


4

Sure, Lagrange interpolation will work, but I think there's more to be said for the specific question... The lowest such polynomial should have degree $\le n-1$ (this is just true in general for interpolating $n$ points). Note that $$Q(m) = mP(m) - 1$$ has roots at $m = 1, 2, \dots, n$, and has degree $\le n$. Thus $$ Q(m) = c(m-1)(m-2)\dots(m-n) $$ for ...


4

As other answers show, there is a large class of solutions to your functional equation if one does not restrict $f$ more. As per your request, we shall try the restriction that $f$ should be a polynomal. Two polynomial solutions jump into our eyes, namely the zero polynomial $f(x)=0$ and $f(x)=2$. Neither of these has $f(4)=65$, though. If $f$ is a ...


3

You can think of $\mathbb{F}_5$ as $\{0,1,2,3,4\}$ or you can think of it as $\{-2,-1,0,1,2\}$. They are both simpler ways of writing the same thing: $$\mathbb{F}_5 = ...


3

$\renewcommand{\phi}{\varphi}\newcommand{\Q}{\mathbb{Q}}$First of all, any such $\phi$ will send each rational number (that is, the constants) to itself. This is because $\Q$ is the prime field here, i.e. the subfield generated by $1$. Now the universal property of polynomial rings tells you that any ring homomorphism $\phi : \Q[x] \to \Q[x]$ which sends ...


3

$$\begin{align} f'(1)&=\sum_{j=1}^{25}j(26-j)\\ &=2\cdot13^2\cdot 5^2-5^2\cdot13\cdot17\\ &=13\cdot5^2(26-17)\\ &=3^2\cdot 5^2\cdot13 \end{align}$$ So the number of divisors is $3\cdot 3\cdot 2=18$. I have used these formulas: $$\sum_{j=1}^n j=\frac{n(n+1)}2$$ $$\sum_{j=1}^n j^2=\frac{n(n+1)(2n+1)}6$$ And the number of divisors of ...


3

We have $b+c = -a$ and $bc=-\frac{q}a$. This means $b$ and $c$ are roots of the quadratic $$y^2+ay-\frac{q}a = 0 \implies ay^2 + a^2y - q = 0$$


3

We have $\alpha+\beta=2$ and $\alpha\beta=\frac{4}{3}$. The first two terms add up to $\frac{\alpha^2+\beta^2}{\alpha\beta}$, which is $\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$. This is $1$. The term $2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$, that is, $2\left(\frac{\alpha+\beta}{\alpha\beta}\right)$, is equal to $3$, and the term ...



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