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 ).

Aug 6, 2016 · How would you show that if $ac≡bc$ $\\mod m$ and $\\gcd(c,m)=d$, then $a≡b$ $\\mod \\frac{m}{d}$? Any help would be much appreciated!

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 …

Prove that (Zn, +) (Z n, +), the integers (mod n) (mod 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 closed- …

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 ...

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 $21$, …

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 …

Proof for: $ (a+b)^ {p} \equiv a^p + b^p \pmod p$ Ask Question Asked 13 years ago Modified 5 years, 8 months ago

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.) …

Do you know that $ (a,m)=1$ gives you some $r$ with the property that $ar\equiv 1 (\mod m)$? Use this by $a^k\equiv b^k$ means $ba^k\equiv b^ {k+1}\equiv a^ {k+1 ...