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 …
modular arithmetic - Prove that $a^ {2^n} \equiv 1 \pmod {2^ {n+2 ...
Apr 1, 2020 · Let's prove that for any odd integer ( a ), ( a^ {2^n} \equiv 1 \pmod {2^n} ) for all ( n \geq 3 ). First, we observe that ( a ) is odd, which means ( a = 2k + 1 ) for some integer ( k ).
How to solve $a^7 \\equiv a \\pmod {42}$ involving congruences?
For all integers $a$ prove that $$a^7 \equiv a \pmod {42}.$$ There is no use telling you all what and how much I tried because I cannot even understand the problem ...
number theory - Proof for: $ (a+b)^ {p} \equiv a^p + b^p \pmod p ...
Proof for: $ (a+b)^ {p} \equiv a^p + b^p \pmod p$ Ask Question Asked 13 years ago Modified 5 years, 8 months ago
What is the difference between $x \\bmod y$ and $x \\pmod y$?
Jul 4, 2015 · @AlanU.Kennington \pmod stands for parenthesized mod, not all that difficult to remember either if you take in consideration that $\TeX$ is all about formatting :) Note that "remainder" and …
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 $21$, …
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 …
Solve $x^3 \equiv 1 \pmod p$ for $x$ - Mathematics Stack Exchange
Dec 28, 2010 · How can I find solution for $x^3 \\equiv 1 \\pmod p$ ($p$ a prime) efficiently? Trivial root is $x_1 = 1$. I need to find other roots $x_2, x_3$.
Prove that for any positive integer $a,$ $a^{561} \\equiv a \\pmod{561}.$
Jun 15, 2020 · Prove that for any positive integer a a, a561 ≡ a (mod 561) a 561 ≡ a (mod 561). (Hence, 561 561 is a pseudoprime with respect to any base. Such a number is called a Carmichael number.) …
Show that if $a \\equiv b \\pmod n$, $\\gcd(a,n)=\\gcd(b,n)$
If $\rm\ d\ |\ n\ |\ a\!-\!b\ $ then $\rm\ d\ |\ a \!\iff\! d\ |\ b, \ $ i.e. $\rm\bmod d\!:\ $ if $\rm\ a\equiv b\ $ then $\rm\ a\equiv 0\!\iff\! b\equiv 0$ So $\rm\, \ {n,a\},\ \ {n,b\}\ $ have the same common divisors …