Quote from: vjudeu on March 21, 2024, 05:49:54 AM

The question is: does it mean that there is some kind of connection between y^2=x^3+7, and for example y^2=x^3+2? Or maybe there is another connection, where points on curves with identical p-value and n-value can be mapped? Does it mean, that if we have b=0x7, where there are "n" points, and for example b=0xc curve also has the same amount of points, then does it mean we can map them 1:1?

For y²=x³+d (mod p), and d being non-zero integer, the group falls into one of these different sets:
1: 2x 2 * 3 * 20412485227 * 83380711482738671590122559 * 5669387787833452836421905244327672652059 2: 3x 3 * 13² * 3319 * 22639 * 1013176677300131846900870239606035638738100997248092069256697437031 3: 109903 * 12977017 * 383229727 * 211853322379233867315890044223858703031485253961775684523 4: 3 * 199 * 18979 * 5128356331187950431517 * 1992751017769525324118900703535975744264170999967 6: 14x 2 * 7 * 10903 * 5290657 * 10833080827 * 22921299619447 * 41245443549316649091297836755593555342121 7: 115792089237316195423570985008687907852837564279074904382605163141518161494337
Here 'x' is the torsion group - then the group is noncyclic (as a whole).

One can move between different d by multiplying the whole equation by k⁶, and getting the new coordinates with the new d (the usual isomorphism):
y² = x³ + d k⁶*y² = k⁶*x³ + k⁶*d (k³*y)² = (k²*x)³ + k⁶*d

Trying to move between these six groups doesn't work - either k³ and/or k² are outside the usual group of numbers mod p.

If one thinks a bit it is obvious. Groups have different number of points, so - when trying to map - every point from one group corresponds to all the points in another.

So, only thing needing in order to jump to an isomorphic equation is taking sixth root of some number a (mod p).
a^p = a a^p+1 = a² a^(p+1)/4 = a^1/2 a^p+2 = a³ a^(p+2)/9 = a^1/3 (a^(p+1)/4)^(p+2)/9 = a^1/6 = k
One should then check if this root exists, i.e. if k⁶=a

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