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21

I will prove the following fact: Let $x$ be a real number such that $x^n$ and $(x+1)^n$ are rational. Then $x$ is rational. Without loss of generality, I assume that $x \neq 0$ and $n>1$. Let $F = \mathbb Q(x, \zeta)$ be the subfield of $\mathbb C$ generated by $x$ and a primitive $n$-th root of unity $\zeta$. It is not difficult to see that ...


20

Here is a proof which does not require Galois theory. Write $$f(z)=z^n-x^n\quad\hbox{and}\quad g(z)=(z+1)^n-(x+1)^n\ .$$ It is clear that these polynomials have rational coefficients and that $x$ is a root of each; therefore each is a multiple of the minimal polynomial of $x$. If $x$ is irrational then its minimal polynomial has degree at least $2$, and so ...


16

Consider the identity $(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$ Differentiating both sides w.r.t. $x$, we get, $x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$ Now, the equation becomes, $(x^4-4x-1)(x^3-1)=0$ $\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity) or $x^4-4x-1=0$ $\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$ ...


11

$$2x^4+3x^3-32x^2-48x$$ $$=x(2x^3+3x^2-32x-48)$$ $$=x(x^2(2x+3)-16(2x+3))$$ $$=x(2x+3)(x^2-16)$$ $$=x(2x+3)(x-4)(x+4)$$


6

Hint: Let $P_n(x)$ be your polynomial. Then show $P_{n+1}(x)=P_n(x-\sqrt{n+1})P_n(x+\sqrt{n+1})$, and show inductively that $P_n(x)$ always has only integer coefficients.


5

let $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$ then we have $$\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}$$ use this identity $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$ so $$\begin{cases} ...


5

We can in fact give a stronger result: Let $P_N$ denote the set of monic polynomials of degree $n > 0$ in $\mathbb{Z}[x]$ whose coefficients all have absolute value $< N$. S. D. Cohen gave in The distribution of Galois groups of integral polynomials (Illinois J. of Math., 23 (1979), pp. 135-152) asymptotic bounds for the ratio in the above limit, and ...


5

Since $P_k(x)-P_k(x-1)=x^k$, $$P_k'(x) - P_k'(x-1) = kx^{k-1} = k(P_{k-1}(x)-P_{k-1}(x-1))$$ Hence, integrating from $0$ to $x$, we find $$P_k(x)-P_k(x-1) - P_k(-1) = I_k(x) - I_k(x-1)$$ where $I_k(x) = k\int_0^x P_{k-1}(t) dt$. Both $P_k(x)$ and $I_k(x)$ are polynomials. Let $c_k = P_k(-1)$. Then we can rewrite the above as $$(P_k(x) - c_k x) - ...


5

Elementary solution We exploit the insight that the product of all of the roots (and hence the product of the moduli of the roots) is $1$: Since $p$ has odd degree and real coefficients, it has at least one real root. But substituting gives that $p(-1) = 4$ and $p(1) = 6$, so $p$ has a real root that in particular does not have unit modulus. Thus $p$ must ...


5

Consider $x = \frac{10}{12}$: $$\begin{align*} f\left(\frac{10}{12}\right) &= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\ &= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\ &= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\ &= -\frac{6}{144}c \end{align*}$$ And consider $x=0$: $$f(0) = 0^2a + 0b + c = c$$ If ...


4

Hint; Note that $$f(-1)=f(0)=0\implies f(x)=x(x+1)(mx+b)$$ by the condition: $f(x-1)-f(x)=4x^{2}$, then you get to that, $m=\frac{-4}{3}$ and $b=\frac{-2}{3}$.


4

Well, you can use $f(x-1)=f(x)+(2x)^2$ to compute values for $f(-1), f(-2)$, and $f(-3)$ starting with $f(0)$. Or you could use finite difference to start with $f(x)-f(x-1)=-(2x)^2$.


4

Then $f(x)=g(x)h(x)$ where $\deg g,\deg h< \deg f$. If $f(x)$ is a prime for infinitely many integers $x$, then either $g(x)$ or $h(x)$ is $1$ for infinitely many $x$, so...


4

first I prove right hand: let $$a_{n}=\sqrt{x}-P_{n}(x)$$ then we have $$a_{n+1}(x)=a_{n}(x)\left[1-\dfrac{\sqrt{x}+P_{n}(x)}{2}\right] \le a_{n}(x)\left(1-\dfrac{\sqrt{x}}{2}\right),x\in[0,1]$$ so $$a_{n}(x)\le a_{0}\left(1-\dfrac{\sqrt{x}}{2}\right)^n =\sqrt{x}\left(1-\dfrac{\sqrt{x}}{2}\right)^n$$ note ...


4

Algebraic Geometry Approach We claim $f_1,f_2,f_3$ have no common $($complex$)$ zeros. Suppose otherwise, that there exists a common zero $(u,v)$; then $$0 = f_1 - 2f_2 = (u^2+v^2-2) - 2(uv - 1) = (u-v)^2,$$ so $u-v=0$. Now we have $0 = f_2 = u^2-1$, so $u = \pm1$. But $$0 = f_3 = u^3+5uv^2+1 = 6u^3 + 1 = \pm 6 + 1 \ne0,$$ contradiction. By Corollary ...


4

I assume you mean $f \in \mathbb{C}[x]$, not $f \in \mathbb{C}[x, y]$, because otherwise $V(f)$ is a curve consisting of infinitely many points for any nonconstant polynomial $f$. Now, if $f \in \mathbb{C}[x]$, you have the following: If $f = 0$ then $V(f) = \mathbb{A}^1$. If $f$ is a nonzero constant function, then $V(f)$ is empty. Otherwise, $V(f)$ ...


3

Here is an alternative proof. Let $P_n=Q_n-1$. Then $P_n=\sum_{j=0}^n (2j-1)X^j$. $P'_n(x)=\sum_{j=1}^n j(2j-1)x^{j-1}$ is plainly nonnegative when $x\geq 0$, so $P_n$ is increasing on $[0,+\infty)$. Since $P_n(0)=-1<0$ and ${\lim}_{+\infty}P_n=+\infty$, we see that there is a unique $x_n\geq 0$ such that $P_n(x_n)=0$. Also, we have the identity (see ...


3

Call $ax+b$ the inverse. Force: $$(2x+1)(ax+b) = 1 \implies 2ax^2+(a+2b)x+b = 1.$$ This suggests $b = 1$. And $a$ must satisfy: $$a +2\equiv0 \pmod 4 \quad \mbox{and} \quad 2a \equiv 0 \pmod 4.$$ Notice that $a = 2$ does the job. I leave you to check that $(2x+1)(2x+1) = 1$...


3

Completing square does the job, $$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})=a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})=a((x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2})=a(x+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a})(x+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a})$$ You could also do it with Vietas formulas ...


3

I'll follow a third way. Let $I=(f_1,f_2,f_3)$ and let's consider $\mathbb C[t,x]/I$. This is isomorphic to $$\mathbb C[x,x^{-1}]/(x^{-2}+x^2-2,x^{-3}+5x^2+1).$$ But $(x^{-2}+x^2-2,x^{-3}+5x^2+1)=(x^4-2x^2+1,5x^5+x^3+1)=(1)$ (why?), so $\mathbb C[t,x]/I=0$.


3

Setting $(a,b,c,d) = (0,0,0,0)$ yields $4P(0)^2 = 2P(0)$, i.e. $P(0) = 0$ or $P(0) = \frac{1}{2}$. However, since all the coefficients of $P(x)$ are integers, $P(0)$ must be an integer. Hence, $P(0) = 0$. Then, setting $(a,b,c,d) = (x,0,y,0)$ yields $P(x)P(y) = P(xy)$ for all reals $x,y$. Also, setting $(a,b,c,d) = (0,x,0,y)$ yields $P(x)P(y) = P(-xy)$ ...


3

you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem


3

In characteristic $0$ the theorem you're speaking of is actually an if and only if: Theorem: If $E$ is a splitting field of $f \in k[x]$, where $k$ is a field of characteristic $0$, then $f(x)$ is solvable by radicals if and only if the Galois group of the extension $E/k$ is a solvable group. In positive characteristic it's not an if and only if. The ...


3

Hint: First, prove that $f(x)$ is irreducible over a field $F$ $\iff$ $f(x+c)$ is also irreducible over $F$ for any $c \in F$. Given this result, note that $f(x-1) = x^5 + 5x^4 - 15x^3 + 20x^2 - 30x + 20$.


3

The first, third and fourth constraints give: $$ p(x) = 1+x+ax^3 $$ (just consider the RHS as a Taylor series in $x=0$), hence by plugging in the third constraint we get: $$ p(x) = 1+x-3x^3.$$


3

More generally speaking, consider the cubic polynomial as $$p(x)=a x^3+bx^2+cx+d$$ for which the first and second derivatives are given by $$p'(x)=3a x^2+2bx+c$$ $$p''(x)=6ax+2b $$and now apply the conditions in the order they are given in the post. So,$$p(0)=d=1$$ $$p(1)=a+b+c+d=-1$$ $$p'(0)=c=1$$ $$p''(0)=2b=0$$ So, you have four simple equations to solve ...


3

Since $a$ and $b$ are the solutions of $x^3-2x+c=0$ we have, $a^3-2a+c=0$ & $b^3-2b+c=0$ Subtracting the above two equations, we get, $a^3-b^3-2(a-b)=0$ $\implies (a-b)(a^2+ab+b^2-2)=0$ $\implies a^2+ab+b^2=2$ (Since $a$ and $b$ are distinct) Now, $$a^2+ab+b^2=2$$ $$a^2(2a^2+4ab+3b^2)=3$$ $$b^2(3a^2+4ab+2b^2)=y$$ Squaring ...


3

You have proved that $x^2 + x + 1 = 0 \implies x^3 = 1$. This is a nice fact, but it doesn't contradict the statement that the second equation has real solutions while the first one doesn't. One way to look at this is the factorization $x^3 - 1 = (x-1)(x^2 + x + 1)$. Clearly, the right-hand side vanishes when $x=1$, but that tells us nothing about $x^2 + ...


3

Let's begin by writing out some of the expansions. $$ \begin{align} e^{x^2} &= 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6!} + \cdots \\ \ln(1 + x^2) &= x^2 - \frac{x^4}{2} + \frac{x^6}{3} + \cdots \\ \cos(2x) &= 1 - \frac{4x^2}{2} + \frac{16x^4}{4!} - \frac{64x^6}{6!} + \cdots \\ 2x\sin x &= 2x^2 - \frac{2x^4}{3!} + \frac{2x^6}{5!} + \ldots ...


3

Hint: Start by assuming $$a_1p_1 + a_2p_2 + a_3p_3 = 0$$ for some scalars $a_1, a_2, a_3$. You have to show that in fact $a, b, c$ are all zero. To do this remember that the equation above is an equality of polynomials. So the coefficients of $x^2$ on the left should equal the coefficient of $x^2$ on the right (which is zero!). Same for the coefficient ...



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