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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 $2$ we 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 gives $x\geq2$ 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.


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