query: What does it mean to say $x_{1} mod -x_{2}$ or $-x_{1} mod x_{2}$ when $x > 0$
positive_passages.docid: 100720
positive_passages.text: Negative quotients and their remainders
negative_passages.docid: 124279
negative_passages.text: $g^k$ is a primitive element modulo $m$ iff $\gcd (k,\varphi(m))=1$
negative_passages.docid: 138088
negative_passages.text: Diffie-Hellman key exchange: Why is $(g^{k_1} \bmod n )^{k_2} \bmod n \equiv (g^{k_2} \mod n)^{k_1} \bmod n$
negative_passages.docid: 40663
negative_passages.text: $a_1 a_1', a_2 a_2', ..., a_n a_n'$ is not a reduced modulo system $\mod p$.
negative_passages.docid: 118511
negative_passages.text: What's the syntax for mod in java
negative_passages.docid: 84199
negative_passages.text: What does the $\|$ operation mean in cryptographic notation?
negative_passages.docid: 96237
negative_passages.text: $1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6
negative_passages.docid: 21108
negative_passages.text: Perfect squares and modularly congruency in mod 5
negative_passages.docid: 52542
negative_passages.text: For which odd positive integer $n$ , is it true that $-1$ is not a positive power of $2$ modulo $n$?
negative_passages.docid: 28330
negative_passages.text: Finding x in $a^{x} \bmod  b = c$ when values a,b, and c are known?
negative_passages.docid: 79135
negative_passages.text: Why do we need in RSA the modulus to be product of 2 primes?
negative_passages.docid: 94855
negative_passages.text: Prove if a (mod n) has a multiplicative inverse, then it's unique
negative_passages.docid: 144067
negative_passages.text: calculating $a^b \!\mod c$
negative_passages.docid: 139155
negative_passages.text: What does 2 to the power x mean in set theory
negative_passages.docid: 122780
negative_passages.text: What does # mean in Mathematica?
negative_passages.docid: 75351
negative_passages.text: How to find inverse Modulo?
negative_passages.docid: 12636
negative_passages.text: Product Rule for mod operator: $ ab\bmod p = ((a\bmod p)(b\bmod p))\bmod p$