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The above mentioned work is focused on a hardware implementation (I have this work as a PDF). I'd suggest you to search for: Colin D. Walter. Montgomery Exponentiation Needs no Final Subtractions. Electronics Letters, 35(21):1831{1832, October 1999. Colin D. Walter. Montgomery's Multiplication Technique: How to Make It Smaller and Faster. In C etin K. Koc ...

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On standard way to compute scalar multiplication is to use Double-and-add algorithm: The idea is to take the binary representation of your scalar $b = b_0 ... b_m$in your case $b = 3$ gives $b_0b_1 = 11$. First you initialize your result $Q$ with $0$. Then for each increasing bit index $i$, you set $Q = 2Q$ (computed with the doubling formula) and if ...

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Window size(w) depends on the NIST curve and the algorithm that you are going to use for point multiplication. I suggest you to look to the paper from Brown et al. Software Implementation of the NIST Elliptic Curves Over Prime Fields. They provide the number of operations in algorithms w. r. t. changing window size for some point multiplication algorithms. ...

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If you mean perfectly secure, then YES OTP is indeed perfectly secure. But if you mean practically secure, then NO OTP is not practically secure. The reason is due to the fact that OTP requires usage of perfectly random numbers used as keys on both sides, however in practice using completely random keys is non-trivial, and usage of Pseudo Random Number ...

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Is this secure? Yes, One-Time-Pads (OTPs) can be proven information theoretically secure. For a sketch of what this means and how to do this, please refer to this previous answer by me. Can I actually use modular addition as encryption like it said in Wikipedia? Yes, any group operation can be used to form a pefectly secret encryption scheme ...

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Your idea of addition modulo 10000 is correct. The correctness follows from the fact that $\mathbb{Z}_{10000} = \{0, 1, \dots, 9999\}$ equipped with addition modulo $10000$ forms a (finite) group. Let $m \in \mathbb{Z}_{10000}$ denotes a PIN code. Now choose a uniformly random element $r \in \mathbb{Z}_{10000}$ and define $c = m + r \bmod 10000$. It is ...

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So this feels like a homework question, as such I"m not going to give you the full answer, but yes, yes you can. https://en.wikipedia.org/wiki/Homomorphic_encryption#Unpadded_RSA Is the best starting point I can give without giving away the barn, but essentially rsa is homomorphic, and you can exploit that and repeated calls to the oracle to do what you ...

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The paper states $R$ contains $2^{255}x^{10}-19$, which represents $0$ in $\mathbb Z/(2^{255}-19)$. In other (slightly simplified) words: $$x^{10} \;\;\text{is}\;\; 2^{-255}\cdot19 \text!$$ (Note that this contradicts your comment to the question: $x^{10}$ does not represent $2^{255}$.) To understand this, recall that a value in \mathbb ... 1 This is just a modular multiplication : \begin{align}17^{93} \mod 23 &= (((((((17^1 )^2 17^0 )^2 17^1 )^2 17^1 )^2 17^1 )^2 17^0 )^2 17^1\\ &= (((((17^2)^2 17)^2 17)^2 17)^2)^2 17 \text{\leftarrow$step 3}\\ &= (((((289 \bmod\ 23)^2 17)^2 17)^2 17)^2)^2 17\\ &= ((((13^2 17)^2 17)^2 17)^2)^2 17\\ &= (((((169 \bmod\ 23) 17)^2 17)^2 ... 1 (a)$\mathbb{Z}_8^* = \{1,3,5,7\}$. (b)$x = 3$. (c) Assuming$a \in \mathbb{Z}^*_{15}$:$a^8 \bmod 15 = 1$and$a^9 \bmod 15 = a$. (d)$M = 4$. 1 Consider two numbers$a$and$b$that square to the same value modulo$n$and don't just differ by the sign. $$a^2 \equiv b^2 \pmod n2$$ $$(a-b)(a+b) \equiv 0 \pmod n$$ Neither of the factors on the left is 0 (or equivalently a multiple of$n$), thus each of them must contain one of the prime factors of$n$. Thus you can use$\operatorname{GCD}(a-b, n)\$ ...

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