I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$  I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$  As $\text{deg}(D)=0$ and $\text{sum}(D)=\infty$ then $D$ is the divisor of a function - we want to find this function  I am given the following: \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}  I have already found (in ) that $$D = \left[(2,-4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$  I am then told that we can write   $$[(4,5)]+[(6,3)]=[(2,4)]+[\mathcal{O}]+\text{div}\left(\frac{y+x+2}{x-2}\right)$$  And thus write $$D=[(2,-4)]+\text{div}\left(\frac{y-2x}{x-2}\right)+[(2,4)]+\text{div}\left(\frac{y+x+2}{x-2}\right)-2[\mathcal{O}]$$  However I am struggling to see how we can find the value of $$\text{div}\left(\frac{y+x+2}{x-2}\right)$$ from what we already know  I can find that \begin{align}\left[(4,5)\right]+\left[(6,3)\right] &amp;=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}\left(\frac{y+x+2}{x-2}\right)\\ &amp;=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}(y+x+2)-\text{div}(x-2)\\ &amp;=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}(y+x+2)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)\\ &amp;=\text{div}(y+x+2)-\left[(2,-4)\right]+3\left[\mathcal{O}\right]\\ &amp;\Downarrow\\ \text{div}(y+x+2)&amp;=\left[(4,5)\right]+\left[(6,3)\right]-\left[(2,-4)\right]-3\left[\mathcal{O}\right] \end{align}  but how can I justify this from only knowing \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}    This question comes in 3 parts:   Step 1 -  Step 2 - This one Step 3 -
58301
I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$  I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$  I know that $\text{deg}(D)=1+1+1+1-4=0$ and it is clear that $\text{sum}(D)=\infty$  Therefore $D$ is the divisor of a function - we want to find this function  I am given \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}  I am then told that we can express $D$ as follows:  $$D = \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$  However, when I expand this back out, I get the following:  \begin{align}D &amp;= \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\text{div}(y-2x)-\text{div}(x-2)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\left(\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\right)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(0,0)\right]+2\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right] \end{align}  which clearly isn't equivalent to the original $D$ unless $$\left[(2,4)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right]=-4\left[\mathcal{O}\right]$$  Can anyone either explain what I've done wrong or show me how I can obtain the above equation please    This question comes in three parts:   Step 1 - This one Step 2 -  Step 3 -
6260
I'm trying to work through the exercises in Otto Forster's book on Riemann Surfaces. While most of them seemed not that hard, this one gives me a headache:  Let $X=\mathbb{C}\setminus\{\pm1\}$  and $Y = \mathbb{C}\setminus\{\frac{\pi}{2}+k\pi~|~k\in\mathbb{Z}\}$. Then $\sin: Y\rightarrow X$ is a covering map. This is clear.  Now consider the following curves:  $$ u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$$  and  $$ v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$$  Let $w_1: [0,1]\rightarrow Y$ be the lifting of $u\cdot v$ and $w_2$ be the lifting of $v\cdot u$ with $w_1(0)=w_2(0)=0$.  Show that $w_1(1) = 2\pi$ while $w_2(1) = -2\pi$.  As of now, I don't have many ideas how to solve this. Of course, the inverse sine is a multi-valued function so the two liftings should use different branches. But I don't see how exactly they come into play.  I'm also not sure whether the solution can be done by calculations alone or if there is some more general principle underlying. Any suggestions would be welcome.
24974
Let $F$ be a field, let $p , q$ be two coprime natural numbers and we consider the two polynomials $$ f(X) = \sum_{i = 0}^{p - 1} X^{k i} \qquad \mbox{ and } \qquad g(X) = \sum_{j = 0}^{q - 1} X^{k j} $$ in $F[X]$, where $k \in \mathbb{N}$. If $x$ is an arbitrary element in $F$, how can I show that $\gcd(f(x) , g(x)) = 1$?
10216
Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me.  Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers.  Preludium   In my lecture we have defined the norm $N_{K/\mathbb{Q}}(x)$ for an element $x \in K$ in the usual way. In a following lecture we "extended" this to the norm of an ideal $\mathfrak{a}$ of $\mathcal{O}_K$ by setting $$ \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|$$ and showed that for the principal ideal $(a)$ with $a \in \mathcal{O}_K$ we always have $\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|$. Later, we introduced the notion of a fractional ideal in $K$ being a non-zero, finitely generated $\mathcal{O}_K$-submodule of $K$. And we showed that a non-zero $\mathcal{O}_K$-submodule $\mathfrak{a}$ of $K$ is a fractional ideal iff there is a $c \in \mathcal{O}_K \setminus \{0\} $ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (thus $c\mathfrak{a}$ is an ideal of $\mathcal{O}_K$). We now extended the norm $ \mathfrak{N}$ by defining for each fractional ideal $\mathfrak{a}$ of $K$ the norm $$ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, $$ $c$ being an element of $\mathcal{O}_K \setminus \{0\}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (exists according to 2.).   Question  How can I show that this latter definition is independent of the chosen $c$?  Edit  I changed the definition in (3) from $ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} $ to $\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} $ and added the absolute value everywhere else, too. I think that was a typo in my lecture notes.  Thoughts (Is it better to add them as comments? I was not sure.)   Let $c, c' \in \mathcal{O}_K \setminus \{0\}$ such that $c \mathfrak{a} \subseteq \mathcal{O}_K $ and $c' \mathfrak{a} \subseteq \mathcal{O}_K$. At the moment I am contemplating if and in which way $c$ and $c'$ are related. I was wondering whether one always has $c'=bc$ (or vice versa) for an $b \in \mathcal{O}_K$. Then we had $$ \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}$$ and it remained to show that $[\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}]$ (maybe because $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).
68529
Problems:Determine the number of functions $$f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$$ satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd.   My Attempt: Each integer in domain has $4$ choices and therefore the total number of functions is $f$ is $4^{1999}.$ Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be $2*4^{1998}.$ I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct?  Edit $1$: I tried to proceed further in the following manner:   For a given mapping $f$ let the number of integers assigned to $2000,2001,2002$ amd $2003$ be $p,q,r$ and $s$ respectively, where $p,q,r,s\in \mathbb{Z}^+$. Therefore the sum in question can be written as $f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.$ Clearly the sum will be odd iff $q+s$  is odd. Also note that $p+q+r+s=1999\Rightarrow p+r$ is even. On the other hand if we let $q+s$ as even then $p+r$ must be even. Now the number of values of $p,q,r$ and $s$ for which the above two lemmas hold is the same and therefore if we let $f$ to be the function which maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $q+s$ is even and $g$ to be the function  hich maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $p+q$ is even we will get a bijection.(Maybe!)
60550
This was Problem 1 from the 2019 International Mathematics Olympiad.  Find all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$  that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whenever $x , y$ $\in$ $\mathbb{Z}$.  And this is my progress...  Let $x=y$ then $f(2x) + 2f(y) = f(f(x+y))$ then $f(2x) + 2f(x) = f(f(2x))$ (substitute $x$ with $y$) then $2f(x) + 2f(x) = f(2f(x))$ (from Cauchy's theory $f(nx) = nf(x)$ where $n &gt; 0$) then $4f(x) = 2f(f(x))$ (from Cauchy's theory $f(nx) =nf(x)$ where $n &gt; 0$) then $2f(x) = f(f(x))$ (divide both sides by $2$) then $f(x) = 2x$ which is one of the satisfied fuctions. Let $x = -y$ then $f(2x) + 2f(y) = f(f(x+y))$ then $f(-2y) + 2f(y) = f(f(-y+y))$ (substitute $x$ with $-y$) then $2f(-y) + 2f(y) = f(f(-y+y))$ (from Cauchy's theory $f(nx)=nf(x)$ where $n&gt;0$ ) then $2f(-y) + 2f(y) = f(f(0))$ (from additional inverse property of any integers) then $2f(-y) + 2f(y) = 0$ (from Cauchy's theory $f(0)=0$ ) then $f(-y) = -f(y)$ (from additional inverse property of any integers, since $f(y)$ $\in$ $\mathbb{Z}$) then all functions $f$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$ that satisfy $f(2x) + 2f(y) = f(f(x+y))$ whereas $x,y$ $\in$ $\mathbb{Z}$ are odd.  I can't continue because I'd almost lost my dinner when I solve this, so I decided to take some Clorazepate and sleep. Can you continue or check my progress? Is it right or not? Can you help me please?
31145
In order to show $e^{x+y}=e^{x}e^{y}$ by using Exponential Series, I got the following:   $$e^{x}e^{y}=\Big(\sum_{n=0}^{\infty}{x^n \over n!}\Big)\cdot \Big(\sum_{n=0}^{\infty}{y^n \over n!}\Big)=\sum_{n=0}^{\infty}\sum_{k=0}^n{x^ky^n \over {k!n!}}$$  But, where should I go next to get $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.  Thanks in advance.
95916
I am trying to plot the elliptic curve secp256k1 y^2=x^3+7 in my latex-document.  \begin{center} \begin{tikzpicture}[domain=-4:4, samples at ={-1.769292354238631, -1.76, -1.74, ..., 2.26, 2.35, 2.7, 2.9}]     \draw[-&gt;] (-2.2,0) -- (3.2,0) node[right] {$x$};     \draw[-&gt;] (0,-2.2) -- (0,4.2) node[above] {$y$};     \draw[-&gt;, color=red] plot (\x,{sqrt(\x^3+7)}) node[right] {$y^2=x^3-2x+2$};     \draw[-&gt;, color=red] plot (\x,{-sqrt(\x^3+7)}) node[right] {} \end{tikzpicture} \end{center}   but this gives me a curve that is interrupted at the left. And to be honest, I don't really know what all of these comments mean.  I would be happy if someone could help me, or has a good tutorial that explains how the plotting in Latex works!  Thanks in advance! And all the best.
108373
If $f:E\rightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:S\rightarrow E$), then is the sheaf $\underline{\omega}_{E/S}:=f_*\Omega_{E/S}^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $\underline{\omega}_{E/S}\cong\mathcal{Hom}_{\mathscr{O}_S}(R^1f_*\mathscr{O}_E,\mathscr{O}_S)$, and the first result on elliptic curves in this book is that $R^1f_*\mathscr{O}_E\cong\mathscr{O}_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).  The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $\underline{\omega}_{E/S}$ is invertible, so that an $\mathscr{O}_S$-basis $\omega$ for $\underline{\omega}_{E/S}$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $\mathscr{O}_S$-basis for $\underline{\omega}_{E/S}$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $\mathscr{O}_S$-basis $\omega$ locally on $S$ allows one to regard $(\Omega_{E/S}^1,\omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $\omega$ locally, how are we getting a global section $\omega\in H^0(E,\Omega_{E/S}^1)=H^0(S,\underline{\omega}_{E/S})$ (unless $\underline{\omega}_{E/S}$ really is trivial)?
32668
$$\begin{align*} x&amp;\equiv 1\pmod2\\ x&amp;\equiv 2\pmod3\\ x&amp;\equiv 3\pmod4\\ x&amp; \equiv 4\pmod5\\ x&amp;\equiv 5\pmod6\\ x&amp;\equiv 0\pmod7\\ \end{align*}$$  So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
23482
I am reading up on ECC and having trouble understanding how these are related.   In a finite field, all point operations are taken modulo $p$.  $n$ is the order of the generator $G$ — which apparently is the number of different points on the curve which can be generated by multiplying a scalar with $G$. This is clear. But —   How is this $n$ chosen? Is $G$ chosen so that $n$ is a certain number? Does $n$ effect the security of the curve? Do all the points of the crypto operations with a given $G$ fall in the set of $n$ points for that $G$?    $h$ is the cofactor which is the $\#E(\mathbb F_p)/n$.    What is the significance of this?  $\#E(\mathbb F_p)$ is the number of points of the curve apparently. Why is dividing that with $n$ interesting?
77204
I'm having a rough time understanding the math behind elliptic curves. I want to implement ECDH where user can define a, b, and p parameters of elliptic curve.  How can I calculate generator base point G? Is it any point that satisfies elliptic curve equation $ y^2 \equiv x^3 + ax + b \pmod p$?  Also, how can I calculate order n and cofactor h given the base point?
1716
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.   Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.  Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.  Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.  In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.  Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped  into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.  Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.  Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.  We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
111392
I am having trouble with this function  $$f(x) = e^{2x}(2x-1)+1$$  I found the domain, domain is $\mathbb R$, it is neither even nor odd function.   I found that $f(x)=0  \Leftrightarrow  x=0$, because it is obvious. I am having trouble with finding the sign of this function. That "+1" is making my life miserable. I can easily find the sign of $e^{2x}(2x-1)$, but what do I do with the constant?   Do I find the minimum of $e^{2x}(2x-1)$ and then add constant to it, and then check the sign?   That seems like too much work, considering that this function is actually a numerator of the first derivative of function that appeared in other graphing assignment. There has to be a better way...  The original graphing assignment was $$g(x)=\ln(e^{\frac{2x}{1-e^{2x}}})$$ You can ignore this one when writing the answer. I included it for the sake of context.   EDIT: By finding the "sign" of a function I mean finding the intervals for which $f(x)&gt;0$ and $f(x)&lt;0$.
1323
Consider the expansion $$(1+x+x^2+x^3+x^4)^{496} = a_0+a_1x+\cdots+a_{1984}x^{1984}.$$   $\quad$ (a) Determine the greatest common divisor of the coefficients $a_3,a_8,a_{13},\ldots,a_{1983}$.      $\quad$ (b) Prove that $10^{340} &lt; a_{992} &lt; 10^{347}$.   Is there an easier way to solve this and is there a formula for the multisection for just the sum of the coefficients and not including $x^k$?  I thought of using the Multisection formula to prove (a). That is, $$\sum_{k \equiv r \pmod{m}}a_kx^k = \dfrac{1}{m} \sum_{s=0}^{m-1} \epsilon^{-rs} f(\epsilon^s x)$$ where $\epsilon$ is a primitive $m$th root of unity and $\displaystyle f(x) = (1+x+x^2+x^3+x^4)^{496}$.  Thus we have $r = 3, m = 5$ and so \begin{align*}\sum_{k \equiv 3 \pmod{5}}a_kx^k &amp;= \frac{1}{5} \sum_{s=0}^{4} w^{-3s} (1+w^s x+(w^{s}x)^2+(w^{s} x)^3+(w^{s}x)^4)^{496}\\&amp;=\dfrac{1}{5}\sum_{s=0}^4 w^{-3s}\left(\dfrac{(w^s x)^{5}-1}{w^s x-1}\right)^{496}\\&amp;= \dfrac{1}{5}\sum_{s=0}^4 w^{-3s} \left(\dfrac{x^{5}-1}{w^s x-1}\right)^{496}.\end{align*}
19715
Reading the book by Diophantus, one may be led to consider the curves like: $y^2=x^3+1$, $y^2=x^3-1$, $y^2=x^3-2$, the first two of which are easy (after calculating some eight curves to be solved under some certain conditions, one can directly derive the ranks) to be solved, while the last , although simple enough to be solved by some elementary consideration of factorization of algebraic integers, is at present beyond my ability, as my knowledge about the topic is so far limited to some reading of the book Rational Points On Elliptic Curves, by Silverman and Tate, where he did not investigate the case where the polynomial has no visible rational points. By the theorem of Mordell, one can determine its structure of rational points, if the rank is at hand. So, according to my imagination, if some hints about how to compute ranks of elliptic curves of this kind were offered, it would certainly be appreciated. Thanks in advance.
65299
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=k$ and $x, y, z, k$ are integers.  Prove that $xyz$ is cube of some integer number.    I was wondering about giving a parametrization for the rational points on the elliptic curve in $\mathbb{P}^2(\mathbb{R})$ $$ x^2 y + y^2 z + z^2 x = k xyz$$ but I am not too confident in the topic, so I am asking you to shed some light.  Thank you for your help.