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 that addition is …
Describe all integers $a$ such that $𝑎^ {111} \equiv 1 \pmod {1111 ...
Aug 2, 2025 · Describe all integers $a$ such that $𝑎^ {111}\equiv 1\pmod {1111}$ I know that this means when you raise $𝑎$ to the power of $111$ and divide by $1111$, the remainder is $1$.
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.
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 three, as it …
What does $ a \pmod b$ mean? - Mathematics Stack Exchange
What does $ a \pmod b$ mean? Ask Question Asked 11 years, 8 months ago Modified 11 years, 8 months ago
Under what conditions is $x^x \\equiv c\\pmod p$?
Aug 4, 2020 · This question was proposed in an Elementary number theory textbook I own. The question stated for the conditions needed such that $x^x \\equiv c\\pmod p$, where $p ...
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
Show that if $\gcd (a,3)=1$ then $a^7 \equiv a\pmod {63}$. Why is this ...
Feb 22, 2020 · This embodies a fact we know about congruences: $ka \cong kb \pmod {kc}$ if and only if $a \cong b \pmod {c}$. So the assumption is necessary to prevent the modulus reducing to $21$, …