# Tag Info

## Hot answers tagged polynomials

12

Yes, there is an integral solution with $x,y\neq 0$. The elliptic curve $y^2=x^3+x+4$ has conductor $\delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=\pm265222$.

5

Write :$$f(x)=x^n-k{x^n-1\over x-1}$$ Say it has double root $a$, then $f(x) = g(x)(x-a)^2$ so we have $$g(x)(x-a)^2(x-1)= \underbrace{x^{n+1}-x^n- kx^n+k}_{p(x)}$$ So $a$ is a root of $p(x)$ and $p'(x)$. Since $$p'(x)= (n+1)x^n-n(k+1)x^{n-1}$$ we have $a=0$ or $a={n(k+1)\over n+1}$ only possible double roots. Is this helpful?

5

Let $p+1$, $q+1$, $r+1$ be our roots. Thus, $p$, $q$ and $r$ are positives and by using the Viete's theorem we need to prove that: $$\sum_{cyc}(p+1)(q+1)+\prod_{cyc}(p+1)\geq3\sum_{cyc}(p+1)-5$$ or $$pqr+2(pq+pr+qr)\geq0,$$ which is obvious.

5

If $y$ is an integer root, so $y^3+1$ is divisible by $y$, which gives not so many cases.

5

$|P(x)|$ is differentiable if it has no real single roots. In other words, whenever $P(x_0)=0$, then $P'(x_0)=0$ as well.

4

Since the cubic $$(u+1)^3-a(u+1)^2+b(u+1)-c= u^3 - (a-3) u^2 + (3-2a+b)u - (c+a-b-1)$$ has all three roots positive, the coefficients $a-3$, $3-2a+b$ and $a+c-b-1$ are all positive. Hence $$5+b+c-3a=(c+a-b-1)+2(3-2a+b)>0.$$

4

We start with $p(x)=\sum_{j=0}^n a_j x^j$. Then we rewrite all $x$ into $(x-\tilde{x})+\tilde{x}$ and binomial expand all $[(x-\tilde{x})+\tilde{x}]^j$ and collect to get (1) (after you correct the power of $(x-\tilde{x})$ to $l$). \begin{align*} p(x) &=\sum_{j=0}^n a_j [(x-\tilde{x})+\tilde{x}]^j\\ &=\sum_{j=0}^n\sum_{l=0}^j a_j \binom{j}{l}(x-\...

4

Write $m=x^2-x+1>0$ then from $m\mid 3x-1$ we have $$3x\equiv 1 \pmod m$$ and since $m\mid 9m$ we have also $$9x^2-9x+9\equiv 0\pmod m$$ So $$1-3+9\equiv 0 \pmod m \implies m\mid 7\implies m\in \{1,7\}$$ $x^2-x+1 = 1\implies x\in\{0,1\}$ $x^2-x+1 = 7\implies x\in\{-2,3\}$ Check every $x$ and you are done.

3

My previous attempt having failed due to dependence between the terms (see edit history if you want to laugh and point), I think the best approach is to see this as a Markov process over prefixes of the sum, with two states: zero and non-zero. Let $n = rs$. There are $q^r$ possibilities for each term, of which $(q-1)^r$ are non-zero, distributed evenly ...

3

If $f_a(x)$ is a polynomial in $x$ of degree $n \ge 1$, and $b \ne 0$, then $f_a(x+b) - f_a(x) = f(x+a+b) - f(x+a) - f(x+b) + f(x)$ is a polynomial in $x$ of degree $n-1$. But note that this is also $f_b(x+a) - f_b(x)$. This implies that all the polynomials $f_a(x)$ for $a \ne 0$ have the same degree. Similarly, the leading coefficient of $f_a(x)$ must be ...

3

My approach would be to do a quick inspection: $$P(x)Q(x) =\prod_{i=1}^{50} (x+i)(x-i) = \prod_{i=1}^{50} (x^2-i^2).$$ This immediate tells us two things: 1) We only need to worry about even coefficients 2) Everything will be terms of squares Consider $a_{100}$ it must be $1$ since the only way to get to $x^{100}$ from this requires multiplying the $x$ ...

3

Let $$Q(x)=\sum_{i=1}^{50} b_ix^i.$$ Then $P(x)=Q(-x)$, so $$P(x)=\sum_{i=1}^{50} (-1)^ib_ix^i.$$ $$a_{100}=b_{50}^2=1$$ $$a_{99}=b_{50}b_{49}-b_{49}b_{50}=0$$ $$a_{98}=2b_{48}b_{50}-b_{49}^2$$ $$=\sum_{i\ne j} ij - \left(\sum_i i\right)^2=-\sum_{i=1}^{50} i^2=-42925.$$ $$a_{97}=b_{50}b_{47}-b_{49}b_{48}+b_{48}b_{49}-b_{47}b_{50}=0.$$ Thus we get $1-(-... 3 Note that$x^2 - y^2 - ixy = y^2 \left( (\frac{x}{y})^2 - i(\frac{x}{y}) -1 \right)$, which is simply a quadratic in$(\frac{x}{y})^2$, multiplied by$y^2$. Factorise the quadratic and then multiply$y^2$back in. Do the same with the other factor to finish. 3 If you just want to know whether there are other integer solutions. You can ask a CAS for that. If you go to the online MAGMA calculator and input Q<x> := PolynomialRing(Rationals()); E00 := EllipticCurve(x^3+x+4); Q00 := IntegralPoints(E00); Q00; It will reply [ (0 : 2 : 1), (4128 : 265222 : 1) ] which means there are only two pairs of integer ... 3 Writing $$\sqrt{5x^2+27x+25}=\sqrt{x^2-4}+5\sqrt{x+1}$$ and squaring we get $$4x^2+2x+4=10\sqrt{x+1}\sqrt{x^2-4}$$ dividing by$2we get $$2x^2+x+2=5\sqrt{x+1}\sqrt{x^2-4}$$ squaring again we obtain $$4x^4+4x^3+9x^2+4x+4=25(x+1)(x^2-4)$$ expanding and combining like terms $$4x^4-21x^3-16x^2+104x+104=0$$ This can be factorized as \left( 4\,{x}^{2}-13\,x-26 ... 3 Yes it is irreducible for all n satisfying |n| > 2. If it were not, then y^3+ny+1 would have an integral root y_0. But this is impossible, as y^3_0 is a multiple of y and ny_0+1 is not for every integer y_0 \not = \pm 1. But for y_0 = \pm 1 note that |n| must be no larger than 2. 3 By the Rational Root Theorem, the only possible rational roots are \pm 1, so that (\pm 1)^3 + n (\pm 1) + 1 = 0. Rearranging gives \mp(n + 1) = 1, leaving at most two values of n for which the polynomial is reducible. 3 The statement is false. Let n = 1, m = 2, a = 2. We have to prove that 2^{2^1} + 1 = 5 divides 2^{2^2} + 1 = 17. Are you sure the +1's aren't supposed to be -1's? Because that theorem \textit{does} hold on first sight. EDIT: Define d := m - n. Now \begin{align*}a^{2^m} - 1 &= a^{2^n \cdot 2^d} - 1 \\ &= (a^{2^n \cdot 2^{(d-1)}})... 2 Partial solution: Note that if x,y \in \mathbb{N} and x^{3} + x + 4 = y^{2} Then x^{3} + x \equiv y^{2} mod(4) y^{2} \equiv r mod(4) with r \in \{0,1\} So, we have that x^{3} + x \equiv 0mod(4) or x^{3} + x \equiv 1 mod(4) Then, the only solutions of this equations is x \in \mathbb{N} such that x \equiv 0 mod(4) Then we have that x = ... 2 Partial answer. Write x^3+x+68 = y^2+64$$so$$(x+2)(x^2-4x+17)=y^2+8^2$$If x=4k+1 then x+2\equiv 3 \pmod 4 so there exists prime p\mid x+2 and p\equiv 3 \pmod 4 which means that p\mid y and p\mid 8 so p=2 which is impossible. If x=4k+3 then y=2n so$$x^3+x+4\equiv_4 2\not{\equiv_4} 0\equiv_4 y^2$$So we must check what happens if ... 2 From the binomial theorem, we know that the sum of the powers of x^3 and a/x^2 must add to 5. So if our term in the expansion is k (x^3)^p (\frac{a}{x^2})^q, p+q = 5. Moreover, we want our term to be constant (no nonzero powers of x), so we have 3p - 2q = 0. Solving these two expressions, we have p = 2, q =3. So we must find a such that the ... 2 Hint:$$(A+B)^5={A}^{5}+5\,{A}^{4}B+10\,{A}^{3}{B}^{2}+10\,{A}^{2}{B}^{3}+5\,A{B}^{4}+ {B}^{5} $$2 Hint: binomial expand (x^3+ax^{-2})^5. 2 From the first equation, x is algebraic. This implies \beta algebraic, which might not hold. There are countably many values of \beta that give solutions. We can enumerate them by considering all triples (k,l,m), finding the roots in x and computing \beta=x^4-x^2-k. 2 As y=x^2-2x=(x-1)^2-1\ge0-1, For two real solutions, we need y_1\ge-1,y_2<-1 for the quadratic equation$$y^2-3y+k+2=0$$Again y_2=3-y_1\le3+1 which is already true k+2=y_1y_2=y_1(3-y_1)=\dfrac94-(y_1-\dfrac32)^2 Finally y_1-\dfrac32\ge-1-\dfrac32 2 We know that the roots of P are \{1, \ldots, 50\} and the roots of Q are \{-1, \ldots, -50\}. Therefore the roots of P \cdot Q are \{-50, \ldots, -2, -1, 1, 2, \ldots, 50\}, which we'll call r_1, \ldots, r_{100}. From the first of Vietas formulas we know that$$ 0 = -50 + \ldots + -2 + -1 + 1 + 2 + \ldots 50 = \sum_{k = 0}^{100} r_{k} = - \frac{... 2 The domain givesx\geq2$and we need to solve $$\sqrt{5x^2+27x+25}=5\sqrt{x+1}+\sqrt{x^2-4}$$ or $$5x^2+27x+25=25(x+1)+10\sqrt{(x+1)(x^2-4)}+x^2-4$$ or $$2x^2+x+2=5\sqrt{(x^2-x-2)(x+2)}$$ or $$\frac{2x^2}{x+2}+1=5\sqrt{\frac{x^2}{x+2}-1}.$$ Now, take $$\sqrt{\frac{x^2}{x+2}-1}=t.$$ Can you end it now? 2 Yes. Write$c(h)$for the content of a polynomial in$\Bbb Z[x]$. Let$m$and$n$be positive integers with$mf$,$ng\in\Bbb Z[x]$. I claim that$c(mnfg)=mn$. Certainly,$mn$divides all coefficients of$(mn)(fg)$but its leading coefficient is$mn$. Then$mn=c(mf)c(ng)$(Gauss's lemma). But$c(mf)\mid m$as its leading coefficient is$m$, and$c(ng)\mid n$... 2 For A, Consider $$B:=\{z\in \Bbb C: f^{(n)}(z)=0\;\text{for some}\; n \in \Bbb N \}=\bigcup_n\{z\in \Bbb C: f^{(n)}(z)=0\}$$ Here$B$is uncountable. That means, atleast one set in the union is uncountable. Thus,$\exists k$so that$\{z: f^{(k)}(z)=0\}$is uncountable, so it has limit point in$\Bbb C$and hence result follows! 2 If$a > 0$,$f(x)$has two sign changes, hence Descartes says there are either$2$or$0$positive real roots (counted by multiplicity). Hint:$f'(x) = 5 x^4 - 5 = 0$for$x = 1$, with$f''(1) = 20 > 0$, so$f$has a local minimum at$x=1$, and the sign of$f(1) = a- 4\$ will determine how many positive roots there are.

Only top voted, non community-wiki answers of a minimum length are eligible