I have tried many elliptic curves $y^2 = x^3 + ax +b$ with no success. I know that for prime modules there exists a minimum number of points the elliptic curve has to have, and I couldn't satisfy this for the smallest primes. So I decided to try luck with modules with few quadratic residues such as 8. But again, no luck.
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1Do you know Hensel's lemma? – Servaes Nov 16 '18 at 18:10
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4There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$. – Angina Seng Nov 16 '18 at 18:14
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If $p | n, p^2 \nmid n$ then let $e_p \equiv 1 \bmod p, e_p \equiv 0 \bmod \frac{n}{p}$, the map $(x,y) \mapsto (e_p x,e_p y)$ is an embedding $E/(\mathbb{Z}/p\mathbb{Z}) \to E/(\mathbb{Z}/n\mathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius. – reuns Nov 16 '18 at 19:30
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@LordSharktheUnknown Why not? Is there a way to write it down then? – SlowerPhoton Nov 16 '18 at 20:31
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@Servaes No, reading it now. – SlowerPhoton Nov 16 '18 at 20:31
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Do you want to require $n>1$? – Somos Nov 16 '18 at 20:55
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I couldn't find the elliptic curve modulo 2 even when I used the general form of equation $y^2+a_1xy + a_5y = x^3 +a_2x^2 + a_3x + a_4$ – SlowerPhoton Nov 17 '18 at 18:57
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The answer is: $y^2 + y = x^3 + x +1 \pmod{2}$. The only element here is the point in infinity: the neutral element.
SlowerPhoton
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You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x \in \mathbf Z/3\mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $\mathbf Z/3 \mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
Alex J Best
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