# Tag Info

5

There is only one solution(except $1=1$). There is a proof in Mordell's book on Diophantine Equations. The problem is attributed to Lucas: with N > 1 is when N = 24 and M = 70. This is known as the cannonball problem, since it can be visualized as the problem of taking a square arrangement of cannonballs on the ground and building a square pyramid ...

4

Not a coincidence, definitely. $70$ is a Pell number, so $2\cdot 70^2+1=99^2$, and some solutions of $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = q^2$$ can be derived by imposing that both $2n+1$ and $\frac{n(n+1)}{6}$ are squares: that leads to a Pell equation.

2

No, such $n$ does not exist in general. Take for instance $E: y^2=x^3+x+2$ defined over $\mathbb{F}_5$. Then, $E(\mathbb{F}_5)$ has four points $$(0 : 1 : 0), (1 : 2 : 1), (1 : 3 : 1), (4 : 0 : 1)$$ and the group $E(\mathbb{F}_5)$ is cyclic of order $4$. Now take $Q=(1:2:1)$, which is a point of order $4$, and $P=2Q=(4 : 0 : 1)$. Then, there is no $n$ such ...

2

The notation $\tfrac{1}{p}P$ is a little bit ambiguous. I understand (b) as saying that there is a point $Q$ such that $pQ=P$ and $Q$ is defined over $K_{\lambda}=L_w$ for any $w \mid \lambda$. This is clearly equivalent to saying that $L_w(Q)=L_w$, i.e. that all $w$ split in $L(Q)$. Because all the $p$-torsion points are defined over $L$, all the points $Q'... 1 A morphism$\phi: \mathbb{P}^m \rightarrow \mathbb{P}^n$can be given by$\phi =(F_0: \dotsc : F_n)$, where the$F_i$are homogeneous polynomials of same degree in$m$variables (See Remark 3.2 on the same chapter of Silverman). The only problem is that the$F_i$might have common zeros in$\mathbb{P}^m$. Say$P$is one such point, then$\phi(P) = (0: \...

1

Well, use part a of the question, where you have proved that $(\phi_q-p^m)^2=0$. Suppose that $\phi_q-p^m\neq0$. Then $\phi_q-p^m$ is surjective on $E(\bar{\mathbf F}_q)$ by Theorem 2.22. Then $(\phi_q-p^m)^2=(\phi_q-p^m)\circ(\phi_q-p^m)$ is surjective too. But this endomorphism was shown to be $0$ on $E(\bar{\mathbf F}_q)$. Hence $E(\bar{\mathbf F}_q)=\{0\}... 1 I had the same question. I've solved the problem partially, so I share my idea. I extend the set$S$to$S' = S \cup \{v \in M_K : E' \mbox{ has bad reduction at } v\}$. I use the notation same as Silverman's AEC. Let$v \in S'$, and consider the localization at$v$. A commutative diagram of exact sequences of$G_{\bar{k_v}/k_v} = G_v/I_v\$-module \begin{...

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