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Why is the cofactor of the twisted Edwards curve equal to 8?

ro flag

While The cofactor of the Edwards curve is chosen $4$ in standards, the cofactor of the twisted Edwards curve is chosen $8$. I can't understand the reason for this. Can we choose cofactor $4$ for the twisted Edwards curve? What happens in this case? Is there any security problem in this case?

tr flag
[Here](https://moderncrypto.org/mail-archive/curves/2016/000788.html) is a good explanation (sorry I can't turn it into a proper answer right now)
mehdi mahdavi oliaiy avatar
ro flag
Thank you very much. That was the answer I was looking for. But I can't understand this sentence. "Bernstein chose {8,4} so that security measures on the curve would automatically protect the twist as well". What is the reason for this? This is not obvious.
tr flag
To avoid small subgroup attacks when doing X25519 Diffie-Hellman, the lower bits of scalar is cleared to make them multiples of 8 (the cofactor). This also works for the twist, since 8 is also a multiple of 4. But it wouldn't work the other way around: if you make then multiple of 4, that won't protect you in the twist (which would required multiplying by 8).
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