# Tag Info

## Hot answers tagged discrete-logarithm

1

First, I think you have a typo in your question since in the original article $s = (M - x y)(r^{-1}) \mod p-1$, and not $s = (M - x^y)(r^{-1}) \mod p-1$. Knowing that, then we can construct $s_2$ from $s, r, M$ and $M_2$: $s_2 = s + (M_2 - M)r^{-1} = (M - x^y)r^{-1} + (M_2 - M)r^{-1} = (M - x^y + M_2 - M)r^{-1} = (M_2 - x y)r^{-1}$ A valid signature for ...

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