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I prove the gcd $\,d =7\,$ if $\, x = 5\!+\!7n,\,$ else $\,d=1.\,$ First $\,x^3\!+\!1 = (x\!+\!1)h(x),\ h(x) = x^2\!-\!x\!+\!1.\,$ Let $\,g(x) = 3x^2\!+3x+1.\,$ Then $\,\color{#c00}{(x\!+\!1,g(x))} = (x\!+\!1,g(-1)) = (x+1,1)= \color{#c00}1.\,$ Therefore $\,d = ((\color{#c00}{x\!+\!1})h ,\color{#c00}g) = (h,g) = (h,\, g\ {\rm mod}\ h) = ... 2 First $$3(x^3+1)-(3x^2+3x+1)(x-1)=2x+4$$ and $$2(3x^2+3x+1)-3(2x+4)(x-1)=14$$ Thus, we have $$(3x^2-6x+5)(3x^2+3x+1)-9(x-1)(x^3+1)=14$$ and thus,$(x^3+1,3x^2+3x+1)\mid14$. Since$3x^2+3x+1=6\binom{x+1}{2}+1$, it is always odd. Thus, we can improve the statement to $$(x^3+1,3x^2+3x+1)\mid7$$ If we look mod$7$, we see that the gcd is$7$when ... 2 With your procedure you found that the GCD between the two polynomials$x^3+1$and$3x^2+3x+1$in$\mathbb{Q}[x]$is$7$, or equivalently$1$, because the GCD of polynomials is defined up to constants (every scalar value$c$divides any polynomial$p(x)\in\mathbb{Q}[x]$). Thus there is not contradiction in your statement. 0 Rewrite$x^5+1$as$x^5+0x^4+0x^3+0x^2+0x+1$so that you have places for all the powers. Then proceed like you would with regular number long division making sure to keep the appropriate places lined up. Note that for this particular problem some of these extra places end up not being used. When you are first learning polynomial long division, it is easier ... 1 x^2 +---------- x^3 + 1 | x^5 + 1 x^5 + x^2 ----------- -x^2 + 1 If you really must have everything line up, you could do 1 +--------------- 1 0 0 1 | 1 0 0 0 0 1 1 0 0 1 0 0 ------------- -1 0 1 where I've omitted the actual monomials and just wrote ... 1 We have$x^5 + 1 = x^2(x^3+1) + (1-x^2)$. Note that$1-x^2$cannot be divided by$x^3+1$anymore since it has lower degree. It follows that$x^5+1$gives$x^2$when divided by$x^3+1$, with remainder$1-x^2$. 1 Do you mean a polynomial$P(x,y)$of total degree$3$such that the curve$P(x,y) = 0$is only in one half-plane? For example,$x y^2 + x - 1 = 0$is only in the first and fourth quadrants, while$ x^2 y + y - 1$is only in the first and second. 3 For a function to not have a$y$-intercept, the number$0$would have to not be in its domain.$0$is the the domain of every polynomial, including every cubic function, so they all have$y$-intercepts. 3 The notion of degree is well-defined for rational functions. In this case, it is a rational function of degree$9$. The degree of a rational function$P(x)/Q(x)$is$\max(\deg P,\deg Q)$(where$P$and$Qare polynomials with no common (nontrivial) factor and Q not identically zero). You can see this because your function can be written as $$f(x)=\frac{x^9 ... 3 An infinitely-differentiable function f(x) defined for all real x is a polynomial of degree n if and only if f^{(n+1)}(x) is identically zero, while f^{(n)}(x) is not. (Here f^{(k)}(x) denotes the k-th derivative of f(x).) 15 That is not a polynomial because of the 1/x. A polynomial in x is of the form a_nx^n+a_{n-1}x^{n-1}...+a_1x+a_0 where a_0,a_1,...,a_{n-1},a_n are constants. 0 On the domain of [-1, 1], x = \cos(\arccos(x)) We can substitude x \over b in place of the original x in order to make a similar equality, x = b\cos(\arccos{x \over b}), valid on [-b, b]. This can be generalized to x^p = b^p\cos(\arccos{x^p \over b^p}) Therefore, my original function, f(x) = x - \frac16 x^3 + \frac1{180} x^5, is equivalent to: ... 2 It is quite likely that the original cubic polynomial is the same, under a linear invertible change of variables with rational integer coefficients, to the one in the Corollary on the final page of Nowlan 1926. The behavior as far as represented primes \pm 1 \pmod 9 and primes giving only trivial zeroes 2,4,5,7 \pmod 9 (as your 13: i am calling this ... 1 OK, the discussion in the comments is getting a little confusing, so let me write a full answer instead. I'll try to spell everything out in detail. We are assuming statement (a) and trying to deduce (b). Of course, we can assume that C_1 \cap C_2 contains at least 4 points. Claim 1: No 3 points of C_1 \cap C_2 lie on a line. Proof of claim: A ... 1 Since P has constant term 0, we have P(0) = 0. The remainder of P(x) after division by x-1821 is just P(1821). We now have P(1821) = P(0) + \sum_{k=0}^{1820} P(k+1) - P(k) = 0 + \sum_{k=0}^{1820} k = \frac{1820 \cdot 1821}{2}. 3 For a more general approach, let \alpha be any algebraic number; suppose that \alpha has minimal polynomial p(x) and you want to evaluate 1/f(\alpha), where f is a polynomial which is not a multiple of p. (If f is a multiple of p then f(\alpha)=0 and so 1/f(\alpha) makes no sense.) Use the Euclidean algorithm for polynomials to divide ... 3 Write \alpha = \sqrt[3]{2}. We have (\alpha-1)(\alpha^2+\alpha+1) = \alpha^3 - 1 = 1, hence \frac{1}{1-\alpha} = -\alpha^2 - \alpha - 1. 2 Try multiplying by$$\frac{2^{2/3}+2^{1/3}+1}{2^{2/3}+2^{1/3}+1}$$2 Supplementing Will's answer with the following suggestion for M. The key is to observe that the element 3 is a cubic root of unity in the field \Bbb{F}_{13}. But there are no ninth roots of unity in that prime field, as 9\nmid (13-1). Therefore the polynomial x^3-3\in\Bbb{F}_{13}[x] is irreducible (all its zeros are clearly of multiplicative order ... 3 If you can find a matrix M of integers such that$$ \det (aI + b M + c M^2) $$is your polynomial, you have a method. For example, with$$ M \; = \; \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) , $$we get$$ aI + b M + c M^2 = \left( \begin{array}{rrr} a ... 2 Let \begin{align} p(t)&=1-t+t^2-t^3+\cdots+t^{2N-2}-t^{2N-1}\\ &=(1-t)(1+t^2+t^4+\cdots+t^{2N-2})\\ &=(1-t){1-t^{2N}\over1-t^2}\\ &={1-t^{2N}\over1+t} \end{align} Clearly\sup_{t\in[0,1]}|p(t)|\le1$and$|a_0|+|a_1|+\cdots=1+1+\cdots=2N$. So whatever$C$is, take$N$so that$C\lt2N$. 1 To justify the very first assertion, note that if$x + 1 | (p')^2$, then$-1$is a root of$(p')^2$; then$-1$is also a root of$p'$, so$x + 1 | p'$. The final part of the argument isn't clear; you say $$h(-1) = (x + 1) k(x), k(x) \in \mathbb{R}$$ but$h(-1)$is a number. For a different approach, note that$0$is the only constant solution. Otherwise, ... 1 Hint: The$0$polynomial obviously works. Now look for non-zero polynomials that satisfy our condition. Take such a polynomial$p$, and suppose that$p$has degree$n$. Then$(x+1)p(x)$has degree$n+1$, and$(p'(x))^2$has degree$2(n-1)$. That gives$n=3$. You observed that$x+1$divides$p$. Indeed$(x+1)^2$divides$p$. Thus$p$has shape ... 1 There is actually a simple reason why this is true. The set$S$of extremal points of a second degree polynomial must be an affine subspace. An affine subspace which is invariant under permutations of the coordinates must contain a point on the form$(p,\ldots,p)$. To see this, assume we have two distrinct points$P=(p_1,\ldots,p_n)$and ... 3 Here is a general criterion that is useful. Let$\rm\,D\,$be a domain with fraction field$\,\rm K.$$$\rm f\,\ is\ prime\ in\ D[x]\iff f\,\ is\ prime (= irreducible)\ in\ K[x]\ and\,\ f\,\ is\ superprimitive$$ $$\rm where\,\ f\,\ is\ {\bf superprimitive}\ in\ D[x]\,\ :=\,\ d\,|\,cf\, \Rightarrow\, d\,|\,c\,\ \ for\ all\,\ c,d\in D^*$$ 0 Using the basis of monomials$x^j$,$j = 0 \ldots n$, of the space${\mathcal P}_n$of polynomials of degree$\le n$, the matrix representing$W$is upper triangular. The eigenvalues of$W$(counted by algebraic multiplicity) are the diagonal elements of this matrix. In particular, the diagonal entry corresponding to basis element$x^n$is$\sum_{k=1}^n ...

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$\;x^2-a\;$ reducible in $\;F\;\implies \sqrt a\in F\;\implies\;\exists\,p(x)\,,\,q(x)\in\Bbb Z_2[x]\;\;s.t.\;\;$ $$\sqrt a=\frac{p(a)}{q(a)}$$ Square now and get a contradiction to the fact that $\;a\;$ is transcendental over $\;\Bbb Z_2\;$

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Do I understand correctly, that $Q^{(k)}_k$ is a $k$-th derivative of $Q_k(x)$? If not, disregard the rest. Let $Q_k(x) = \sum_{i=0}^k q_{k,i}x^i$, then $Q^{(k)}_k = q_{k,k}k!$ and $c = \sum_{k=0}^n \binom{n}{k} q_{k,k} k!$ For polynomial $y = \sum_{i=0}^n y_ix^i$ to satisfy the equation, $y_i$ must solve the system of linear equations I am too lazy to ...

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Hint: Note that you can combine the terms in the denominator, then simplify the whole thing into a single term after that.

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This system is a ordinary system of $n$ polynomials of degree $3$ in $n$ variables written in a fancy form. So it may have up to $3^n$ solutions. The usual techniques apply, available are Gröbner based methods if the matrices have rational entries and homotopy continuation methods like LHCPack. Project-Lift-Intersect methods are becoming available, but are ...

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You just need to check that the resultant $r(t)=Res_x(f(x), g_t(x))$, which is a polynomial in $t$, is not the zero polynomial. Any root of $r(t)$ marks a value of $t$ where both polynomials have a common factor. If $r(t)\ne 0$, the polynomials $f(x)$ and $g_t(X)$ are relatively prime. And as in the other answers, as $r(t)$ only has a finite number of ...

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This may not be exactly the type of reference you're seeking, but G. H. Hardy's book Orders of Infinity may be of interest. Wikipedia's definitions of "degree" are defined by a function's behavior near infinity, and therefore do not bound the number of real roots for an arbitrary function. If $b > 0$ and $c$ are real numbers, the function $$f(x) = c + ... 21 It's like this: x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2 - (\sqrt2x)^2 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1) 16 Group each complex root \alpha with \bar\alpha:$$ (x-\alpha)(x-\bar\alpha)=x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha\in{\Bbb R}[x]. $$3 In complicated terms, the field extension \mathbb C / \mathbb R has degree 2, so you expect every quartic to be reducible over \mathbb R. In simple terms, since you have the four roots of this polynomial, there is a nice way to group them together : the roots e^{i\pi/4} and e^{7 i \pi/4} are conjugate, so they are the roots of the same quadratic ; ... 3 let x_i be the roots of f. g_t and f are not primes iif g_t(x_i) \neq 0 for every x_i. So g_t and f are primes iif t no in the finite set of roots of$$b_{m}x_i^m+(b_{m-1}+t)x_i^{m-1}+\cdots+(b_{1}+t^{m-1})x_i+(b_{0}+t^m)$$2 Here is a series of steps, which seem mostly true to me. Fact: f(x) has at most n distinct roots. Claim: There exists a map G: [0,1] \rightarrow (\alpha_1, \alpha_2, \ldots, \alpha_m) which is differentiable in each coordinate, and  \alpha_i are roots of  g_t(x) (with multiplicity). Possible Proof: Use Inverse/Implicit function theorem till 2 ... 1 In abstract algebra, one distinguishes between polynomials and polynomial functions. A polynomial f in one indeterminate X over a ring R is defined as a formal expression of the form f = a_nX^n + \dots + a_1X^1 + a_0X^0 \tag{1} where n is a natural number, the coefficients a_0, \dots, a_n are elements of R, and X is a formal symbol, whose ... 1 First question: I can't see why the author (and the OP) starts with a Dedekind domain since then A[y] is not necessary factorial (if A=\mathbb Z[\sqrt{-5}] it isn't even a GCD domain) in order to ensure that A[y]/(f) is a domain, that is, f is a prime element of A[y]. However, if A is a UFD, then all is okay and we can move on. Since f is monic ... 3 Three answers and nobody has mentioned the Factor Theorem! 1 Factoring cubics can be tricky. But, if you know a few tricks, then it might not be so bad. One trick for this cubic: there appears to be a pattern in the coefficients: +1,-1,-1,+1, and this sums to zero, so it's logical to look at \pm 1 as a root. More generally, you can use the rational root theorem to impose conditions on any rational root of the ... 2 Answer to your 1st question: if a polynomial is degree n, it can have up to n+1 terms, so having 4 terms in a degree 3 polynomial is nothing out of the ordinary. Answer to your 2nd question: finding a factor of a polynomial of degree 3 or more is usually just guesswork. In fact, it's been proven that it's IMPOSSIBLE to find a formula to find all ... 5$$\begin{align}x^3-x^2-x+1&=x^3-x^2-(x-1)\\&=x^2(x-1)-(x-1) \\&=(x-1)(x^2-1)\\&=(x-1)[(x-1)(x+1)]\\&=(x+1)(x-1)^2\end{align}$$3 A nice basis for that space consists of all monomials in the n variables with total degree d. Wait...what about degree d = 2 and n = 3 variables. Listing the basis elements, I see$$ x^2, y^2, z^2, xy, xz, yz, $$which is only 6 dimensions, but 5 choose 3 is 10. Seems as if there might be a mistake in your formula, or perhaps I'm not ... 0 This question was promptly answered as soon as I cross-posted it on MO http://mathoverflow.net/questions/159173/minimize-norm-of-a-polynomial-around-a-circle-count-the-solutions?noredirect=1#comment407081_159173 0 Hint: What is f(b)^p?{}{}{} 1 Hint: Perhaps a little Fermat can help you... 2 The definition of a separable polynomial in Ash 3.4.2 (which you reference) refers to an irreducible polynomial. Your f(x) is completely reduced to linear factors over \mathbb Q. Each of these factors is separable. And, as Ash explains, a polynomial which is the product of separable factors is itself separable. 3 The definition of separability is that an irreducible polynomial is separable iff it has no repeated roots in a splitting field. A criterion for having no repeated roots is given by the lemma involving the derivative, which you stated. But your f is not irreducible. For an arbitrary (possibly reducible) polynomial, we say that it is separable if all of ... 2 Let us assume that \lvert f(z)\rvert\le 1, for all \lvert z\rvert=1, and show that in such case f has to be equal to z^n. Cauchy Integral formula implies that$$ n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{n!}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt, $$and thus,$$ 1=\frac{1}{2\pi }\int_0^{2\pi} ...

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