Number of Solutions to $x^2 - y^2 \equiv 2^r \ell \pmod {2^q}
Dec 12, 2025 · Start asking to get answers Find the answer to your question by asking. Ask question
How to solve $a^7 \equiv a \pmod {42}$ involving congruences?
$$a^7 \equiv a \pmod {42}.$$ There is no use telling you all what and how much I tried because I cannot even understand the problem itself left alone attempting it.
Prove that $ (\mathbb {Z}_n , +)$, the integers $\pmod {n}$ under ...
Prove that $ (\mathbb {Z}_n , +)$, the integers $\pmod {n}$ under addition, is a group. To show that this is a group, I know I need to show three things (in our text, we do not need to show …
Solutions to $x^n \equiv x \pmod {n}$ for composite $n$.
Nov 27, 2025 · By the Chinese remainder theorem, $x^n\equiv x\pmod n$ if and only if the congruences $x^n\equiv x\pmod {p_j^ {r_j}}$ are simultaneously true for $1\le j\le m$.
congruences - $3x\equiv7\pmod {11}, 5y\equiv9\pmod {11}$. Find …
Oct 10, 2013 · 3x $\equiv$ 7 (mod 11) and 5y $\equiv$ 9 (mod 11) 3x $\equiv$ 18 (mod 11) by adding 11 to 7, then 5y $\equiv$ 20 (mod 11) by add 11 to 9. x $\equiv$ 6 (mod 11) by dividing …
Proving $a\\equiv b \\pmod m$ if $a \\bmod m = b\\bmod m$
Nov 20, 2020 · \pmod{m} will produce the parenthetical version of the mod operator (you don’t need to add parentheses); \bmod will produce the “mod” operator. \mod is the worst of the …
What is the difference between $x \bmod y$ and $x \pmod y$?
Jul 4, 2015 · Usually the parenthesized $\pmod y$ goes at the right of the line, right-justified. So it is a qualifier which tells you which equivalence relation is intended.
What integers have order $6 \pmod {31}$? - Mathematics Stack …
Aug 8, 2013 · What integers have order $6 \pmod {31}$? Ask Question Asked 12 years, 4 months ago Modified 12 years, 4 months ago
Assume $p \\equiv3 \\pmod 4$ and $n \\equiv x^2\\pmod p
It cause me some confusion since, for example, $15\equiv 1 \pmod 2$ is so because $ (15-1)=7$ with remainder zero, so I assumed, incorrectly as it seems, that the number on the left side of …
Show that $ka \equiv kb \pmod {n}$ implies $a \equiv b \pmod …
Mar 19, 2024 · Show that $ka \equiv kb \pmod {n}$ implies $a \equiv b \pmod {n}$ if and only if $\gcd (k,n) = 1$. This ambiguous assertion can be interpreted in two different ways: