# Tag Info

22

Yes. In fact, you don't even need the quadratic term: there are degree-$1$ polynomials giving arbitrarily long sequences of primes by the Green-Tao theorem. (Of course, you could also let $P$ be a constant prime).

15

This follows by the Green-Tao theorem: there exist arbitrarily long arithmetic sequences of primes. If $$ak+b,a(k+1)+b,\ldots,a(k+m^2)+b$$ are prime, we can take $P(n)=a(k+n^2)+b$, being prime for $n=0,\ldots,m$. Note that the same works if we require $\deg P=1$.

15

The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $p(x) = -(x + 1) q(x)$ for some quintic $q$. Substituting $x = -1$ gives that $-1$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic ...

7

Quoting from Locating the zeros of partial sums of $\exp(z)$ with Riemann-Hilbert methods: We denote by $p_{n}(z) := 1 + z + \cdots + \frac{z^{n}}{n!}$ the partial sums of the exponential series. The problem to describe the asymptotic distribution of the zeros of $p_n$ was posed and solved in the classical paper of Szegő [11]. He proved that the ...

5

As advised by @πr8, two of the Viète formulas allow to replace $c$ by $\dfrac{1}{ab}$ and $d$ by $-2a-b-c=-2a-b-\dfrac{1}{ab}$. Therefore, the other two Viète formulas can be expressed with variables $a$ and $b$ only, under the form $$f(a,b)=-1 - 2\,a^2\,b - a\,b^2 - 2\,a^4\,b^2 - a^2\,b^3 - 2\,a^3\,b^3 - a^2\,b^4=0 \ \ \ (1)$$ and $$g(a,b)=-a - b + ... 5 If a degree 4 polynomial has 4 real roots, then it must have at least 3 local extremes, so its derivative must have 3 real roots; by repeating this argument, its second derivative must have 2 real roots. But in fact, the second derivative of this function is$$ 12((x-1)^2+1), $$which is clearly positive everywhere. Edit (thank you almagest): This argument ... 5 We're in luck because this is a very small field. We can just test out all the values. Obviously 0 is not a root. f(1) = 1 + 3 + 5 = 9 = 2 \neq 0. f(2) = 8 + 6 + 5 = 19 = 5 \neq 0. f(3) = 27 + 9 + 5 = 41 = 6 \neq 0. f(4) = 64 + 12 + 5 = 81 = 4 \neq 0 f(5) = 125 + 15 + 5 = 145 = 5 \neq 0 f(6) = 216 + 18 + 5 = 239 = 1 \neq 0. So there are ... 5 Map x+(x^2+x)\mapsto -x+(x^2-x) 4 The set of polynomials of even degree is not closed under addition, hence it's not a subspace. But the set of polynomials with only monomials of even degree is a subspace of K[X] since it is K[X^2]. 4 Here's an outline of how I got to this, rather than a pretty, cleaned up proof. First I worked out that$$ (-\log{f})'' = \frac{f'^2}{f^2}-\frac{f''}{f}, $$which I thought might be useful because if you divide by f^2 \geqslant 0, you find terms that look like f'/f. Therefore the inequality is equivalent to showing$$ n (-\log{f})'' \geqslant ...

4

(Edit after a bit more research:) Another widespread notation for $R[\Bbb Z]$ is $R\Bbb Z$. The latter may be preferable if one wants to avoid confusion with the polynomial ring in either a single variable $\Bbb Z$ or the polynomial ring in countably many variables (that happen to be integers, but cannot be added or multiplied as such); then again, for some ...

4

Note $x^3=2$ in the quotient, so $x\cdot x^2=2$ (in the quotient) so $x\cdot {1\over 2}x^2=1$.

3

Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly). Module[{roots}, roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x]; Transpose[{ N[x /. roots], FullSimplify[Element[x, Reals] /. roots], x /. ...

3

It is known that $$x^2+1+x^5=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\tag1$$ Comparing the constants of each side, we have $-1=x_1x_2x_3x_4x_5$. Now, puting in $-x$ in $\text{(1)}$, we have $$x^2+1-x^5=-(x+x_1)(x+x_2)(x+x_3)(x+x_4)(x+x_5)\tag 2$$Multipling $\text{(1)}$ and $\text{(2)}$, we have ...

3

To elaborate (slightly) on the comments: if $P(x)$ were such a polynomial, then both $P(x)+1$ and $P(x)-1$ would have infinitely many zeroes (and hence would be constant). $cos(\pi x)$ is an entire example.

3

You can use Vieta's relations: the sum of the roots (in $\mathbf C$) is $\frac 52$. Now the sum of the found roots is already $\frac 52$. So if there were other rational roots, they would be opposite. Now the equation can be written as $$2x^4+7x^2+15=5x(x^2+5),$$ and having opposite roots would imply they're roots of $x^2+5$, which has no real root.

3

The degree of $q$ is also $n$ (which is even), and the coefficient of the largest exponant is $1$. Thus $q(x)$ goes to $+\infty$ when $x$ approaches $+\infty$ or $-\infty$. Therefore $q$ has a lower bound. Let us denote $m = \inf \limits_{x \in \mathbb{R}} q(x)$. You can easily prove that there exists $a \in \mathbb{R}$ such that $q(a) = m$. We have $q'(a) ... 2 This is probably not the best answer, but this is how I would have done it, since I guess in this kind of exam time mathers? First let$f(x) = x^4 -4x^3 +12x^2 + x -1$then I would noticed that$f(0) = -1$. After that I would look at$f'$We have that$f'(x) = 4x^3 - 12x^2 + 24x + 1$, then one can notice that $$4x^3 - 12x^2 + 24x > 0 \ \forall x > ... 2 Hint: for y=8x^3-6x+1 and y'=24x^2-6 . So ve have two stationary points for x=\pm1/2. Now you can see that there is a positive valued max at x=-1/2 and a negative valued min at x=1/2 and the function is negative for x=-1 and positive for x=1. Use continuity (intermediate value theorem) to show that there are three roots in [-1,1] 2 \cos 3 y=4\cos^3 y-3\cos y. When \cos 3 y=-1/2 and x=\cos y, we have 8 x^3-6 x +1=0, with 3 solutions x=\cos [2\pi(1+3 n)/9] for n\in \{0,1,2\}. 2 Hints. If a=x\bmod (x^4+x+1), then try a^{-1} for x^4+x^3+1, and a^{-1}+1 for x^4+x^3+x^2+x+1. 2 Hagen's answer is great, if you prefer an alternative note that the Chinese Remainder theorem gives$$\Bbb Q[x]/(x(x+1))\cong \Bbb Q\oplus\Bbb Q\cong \Bbb Q[x]/(x(x-1))$$as rings since (x), (x+1), (x-1) are prime ideals. Edit The generalized CRT says that if you have an ideal I which is the product of coprime prime-power ideals i.e. ... 2 Sketch: Suppose that P(a)=P(b)=P(c)=0 where a,b,c are integers. Suppose that there exists a d such that P(d)=1. Prove that (d-a)\mid P(d)-P(a)=1. Hint: Prove (d-a)\mid (d^k-a^k) by factoring. This gives three distinct divisors of 1, do you detect a contradiction? 2 Much of this would be covered in "basic ring theory". Let's say that R is a commutative ring with unity (although more general settings can be accommodated, it seems like a good place to start). As you studied "polynomials" in calculus you were asked to consider them as functions of one or more variables. In abstract algebra there is more of an emphasis ... 2 |f(z)|\leq|z|^n implies f(0)=0. Writing f(z)=z^m g(z) with g(0)\ne0 implies m \ge n and so |z^{m-n}g(z)| \le 1. Now apply Liouville's theorem. 2 You can use the Rational Roots Test: Lemma Let P(X)=a_nX^n+...+a_1x+a_0 be a polynomial with integer coefficient and m,n integers. If x =\frac{m}{k} is a root of P(X) then$$m|a_0 \,;\, k|a_n$$The proof is a pretty simple divisibility problem. For your exercise, the Lemma tells you that m \in \{ \pm 1, \pm 3, \pm 5 \pm 15 \} and k \in \{1, ... 2 The simplest function I can find seems f(x) = \dfrac{2x+2}{x+2}. It is easily verified this satisfies the double inequality for x \ge \sqrt2. This was obtained by looking at linear approximants, and then setting the conditions f(\sqrt2) = \sqrt2, f(x) \ge \sqrt2 and finally the RHS inequality. Even among linear functions, there are of course many ... 2 If you allow non-monic polynomials the question is not very arithmetic in nature. Any class of integer functions closed under the operations f(n) \to Af(n)+B with integer A,B, is guaranteed to have this property for any set that contains long arithmetic progressions (such as the primes, by the Green/Tao theorem as people have mentioned). This ... 2 In general there is no suitable value for c. This has been shown at the answers to this MSE question, with the example$$ f(x) = x^3 + x + 1\in \mathbb{Z}[x].$$2 Hint: For a polynomial of$n$.degree and a coefficient$a_{n-1}\neq 0$. You can always reduce the$(n-1)$-power by$z=x+\frac{a_{n-1}}{n}$. As Dietrich Burde suggested$c=1\$ for this case.

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