Yong Hao Ng
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 Nov28 awarded Popular Question Sep24 awarded Autobiographer Jul2 awarded Curious May20 awarded Yearling Mar19 comment Let $E:y^2 = x^3 + 1$ be an elliptic curve. For each prime $5 \leq p \leq 13$, describe the group $E(\mathbb{F}_p)$. I suppose the first question is whether you need to find all the points by bruteforcing $\mathbb F_p\times \mathbb F_p$. If not, you will need to know that you have indeed found all of them. If you do not have access to Hasse's theorem, this seems quite hard to do. If you do bruteforce, at that point you are basically reduced to a group theory problem and it seems natural to just continue with group theoretic computations. You might want to also note that if $(x,y)\in E(\mathbb F_p)$ then $(x, p-y)\in E(\mathbb F_p)$. Mar19 comment How to show there exists no solution to a discrete logarithm problem on an Elliptic Curve? As for the comment from Álvaro, the idea: you can check if $(1,2)$ and $(4,5)$ are torsion points (Nagell-lutz theorem). If both are torsion, then there are only finitely many points to check. Otherwise, $(1,2)$ is non torsion and you can investigate $[k](1,2)= (x_k,y_k)$. The point is that for sufficiently large $k$, we have $$\max\{x_{k+1},y_{k+1}\}> \max\{x_k,y_k\}$$ so the coordinates is always growing. You can find a $k$ such that in addition $\max\{x_k,y_k\}> \max\{4,5\}=5$. Then for $i\geq k$ we cannot have $[i](1,2)=(4,5)$ by comparing the values. Finally, check for $1\leq i\leq k$. Mar19 comment How to show there exists no solution to a discrete logarithm problem on an Elliptic Curve? In case it is still not clear: for $E/\mathbb Q$, we have discriminant $\Delta$. For each prime $p\nmid\; \Delta$, you have a reduction mod $p$ by reducing each point $P=(x,y)$ to $$\tilde P=(x,y)\pmod p=(\tilde x,\tilde y)$$ (and $E$ too). So if indeed $[k](1,2)= (4,5)$, you have $[k](\tilde 1,\tilde 2)=(\tilde 4,\tilde 5)$ in $\mathbb F_p$ (same addition formula). So it suffices to generate $\langle \tilde P\rangle$ over $\mathbb F_p$ and check if any element matches $(\tilde 4,\tilde 5)$. You can do this for infinitely many $p$ and chances are one of them works if $[k](1,2)\neq (4,5)$. Mar17 awarded Critic Mar12 revised How large do my $2$ primes need to be to “guarantee” a longevity of security for my RSA-encrypted plaintext? Included information for estimating difficulty. Mar12 revised How large do my $2$ primes need to be to “guarantee” a longevity of security for my RSA-encrypted plaintext? Included information for estimating difficulty. Mar12 answered How large do my $2$ primes need to be to “guarantee” a longevity of security for my RSA-encrypted plaintext? Mar12 answered Is it possible to do elliptic curve cryptography over $\mathbb{Q}$ instead of a finite field? Mar6 comment Four integers that satisfy $a+b+c+d\; =\; -3$ and $a^{3}+b^{3}+c^{3}+d^{3}\; =\; 3$ Regarding the insight part: generally integers in 3 variables (your case) means finding integer points on surfaces (Sometimes you can simplify the equations, which is what you should always try first). This is quite difficult if you want the full solution; you need to know some Arithmetic Geometry. For some examples and references you can look up on "rational points on cubic surfaces" (or integral points). If you just want to find some solutions, you can try solving in $\pmod p$ for many primes and combining solutions in Chinese remainder theorem. This helps to exhaust the small solutions. Mar6 comment does this equation has an answer? You need to specify what type of values $a,b$ and $x$ are in. For example, can $a,b$ or $x$ be complex-valued? The tag (diophantine-equations) suggest that $a,b$ and $x$ are integers but you may want to state it clearly in your question. Mar6 comment Why can't the Alpertron solve this Pell-like equation? You may want to also try this Pell equation solver, where the author gave a useful reference regarding the method. Feb28 comment For which values of n does G have an eulerian trail No problems~ No worries, confidence will come with experience. =D Feb28 comment For which values of n does G have an eulerian trail So you are done: Eulerian trail if and only if $n=4$. Feb28 comment For which values of n does G have an eulerian trail I think you meant $4,5,2n-4,2n-3$ has odd degrees (instead of $2n-3,2n-2$, since $2n-2$ has even degrees). That is right. So for $n\geq 5$, these 4 elements are distinct and therefore there cannot be an Eulerian trail. So what you said is right. For $n=4$ they merge into $2$ so you have the Eulerian trail. You have listed one of them: $1,4,7,2,5,8,3$. For $n=3$ the graph is not connected so it does not matter. Feb27 comment For which values of n does G have an eulerian trail @Studentmath No problem, you can ping me in chat if you would like to talk to me. =D Feb27 comment For which values of n does G have an eulerian trail @Studentmath You are welcome! Do you have access to textbooks? It should be a standard theorem in one of the chapters.