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bc on unix: obase=16 ibase=16 p=F401F9E76A0E65D80AA8CF0D526D8D8747E53A3E1223B143AA73F675708ED966AB96965040907CCDF3D5C77904AA0906A6941E3A9C69AEC1F99E73E6EDB07191 q=E29F25EC241F0FEDAD28B8DD1DCBABBD066F4F557467AE6A2CE4ED34F9D93257E2F8C8B6EE1F7A687E386BFEE9C20C3388385E82AFA498237FF801D283216D4D ...


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There is of course but because of carry bits there will be data-dependent "nonlinear" terms. If I do it for 2 bits you can get the idea. It gets unwieldy but you can easily write code to do it for longer bitlengths. The list below is an XOR table expressed as integers: $$ \begin{array}{c|ccc} \oplus & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & ...


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Here's a justification for the estimate in CodesinChaos' comment. Helmut Hasse proved in his 1936 series of papers "Zur Theorie der abstrakten elliptischen Funktionenkörper" that any elliptic curve $E$ over a finite field $\mathbb F_q$ satisfies the inequality $$ \lvert q+1-\#E(\mathbb F_q)\rvert\leq2\sqrt q \text, $$ that is, the number of points is ...


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No, it wouldn't. Very simply put: If you can compute square roots wrt. a composite modulus, then you can calculate other fixed roots, too. This is based on the fact, that being able to compute square roots allows you to factor the modulus: Choose random $x$. Compute $x^2$. apply your square root algorithm to $x^2$. The result $y$ will be a different ...


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OBVIOUS, if you write $f(x)=(1+2^8+2^{16}+2^{24})\times x=K\times x$. Then $f(x)\times f(y)=K^2\times x \times y$. Then $\forall y$, choosing $z \in (\mathbb{Z}/n.\mathbb{Z})^{*}$ allows to dermine the unique $w \in \mathbb{Z}/n.\mathbb{Z}$ satisfying this relation in the multiplicative group.



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