Number of Solutions to $x^2 - y^2 \\equiv n \\pmod {m}
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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.
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 $1\le j\le m$.
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 …
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 …
discrete mathematics - Let $a$ and $b$ be integers and $m$ be a ...
Jun 28, 2022 · Let $a$ and $b$ be integers and $m$ be a positive integer. Prove that $ab \equiv [ (a\pmod m)\cdot (b\pmod m)] \pmod m$ [duplicate] Ask Question Asked 3 years, 5 months …
What is the difference between $x \\bmod y$ and $x \\pmod y$?
Jul 4, 2015 · I'm currently taking Calculus I, so I'm new with mathematical notation. I always thought x mod 3 means the remainder when you divide x by 3. Am I wrong in thinking this is …
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
elementary number theory - Solutions of $x^2 ≡ 1 \pmod {p^2 ...
Oct 20, 2017 · Hint: First note the congruence equation $\; x^2\equiv 1\mod p$ has $1,-1$ as solutions, and that a solution $\bmod p^2$ yields a solution $\bmod p$ by the canonical …
Proof that if $n$ has a primitive root, $x^k \equiv a \pmod n$ has …
Aug 28, 2024 · Then if $a=g^l$ the problem is to find integer $t$ such that $$ (g^t)^k \equiv g^l \pmod n$$ this means $$g^ {kt-l}\equiv 1 \pmod n$$ and this is equivalent to $m |kt-l$.