Jan 3, 2026 · Your solution is fine from what I see, a solution that doesn't use nearby results may sometimes be preferred for additional results obtained along the way or introducing specific …

We claim that in fact $\Phi_n (a) \equiv 0 \pmod {p}$. Note that this follows from the factorization $\Phi_n (x) = \Phi_r (x)^ {p^m - p^ {m-1}}$ over $\mathbb {F}_p [x]$, which we prove here.

Dec 12, 2025 · Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and …

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 $1$.

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 …

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 ...

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 does $ a \pmod b$ mean? Ask Question Asked 11 years, 8 months ago Modified 11 years, 8 months ago

Aug 8, 2013 · What integers have order $6 \pmod {31}$? Ask Question Asked 12 years, 4 months ago Modified 12 years, 4 months ago

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 …