Mathematics

Questions tagged [lifting-the-exponent]

For questions involving LTE = lifting-the-exponent and similar Hensel-like lifting methods

9 questions
8
votes
3 answers

Prove that if $a^p-b^p$ is divisible by $p$, then it is also divisible by $p^2$

$a$ and $b$ are natural numbers and $p$ is a prime number. Prove that if $a^{p}-b^{p}$ is divisible by $p$, then it is also divisible by $p^{2}$. My attempt: Based on Fermat's theorem $(a^{p}-a)$ and $(b^{p}-b)$ are divisible by p, therefore their…
max
  • 324
5
votes
1 answer

Using induction to show that $(\mathbb{Z}/p^a\mathbb{Z})^{\times}$ is cyclic.

I need to show that the group $(\mathbb{Z}/p^a\mathbb{Z})^{\times}$ is cyclic for odd prime $p$ and for $a\in \mathbb{N}^+$. I have already shown that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ has a generator $g$ and so is cyclic. I have also shown that…
Chloe.H
  • 427
4
votes
3 answers

How to prove that if $a\equiv b \pmod{kn}$ then $a^k\equiv b^k \pmod{k^2n}$

What I have done is this: $a\equiv b \pmod{2n}$, $a=b+c\times2n$, for some $c$, $a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$, $a^2-b^2=(b\times c+c^2n)\times4n$, then $a^2\equiv b^2\pmod{2^2n}$. I think that this is right: what I DON’T understand is…
gurghet
  • 529
3
votes
3 answers

Given that $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7$, find the number of positive integer divisors of $n$.

I suddenly recalled one hard question (to me) in a math contest I participated in before. Fortunately I still completely remembered its context as follows: Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by…
user808951
2
votes
0 answers

A divisibility property in a sequence with exponential terms

Given the sequence $(a_n)_{n\geq 1}$ with $a_n = \frac{{2 \cdot 2^{2^{n}} + 1}}{3}$ , prove that $3^n \mid a_{3^n}$ for every $n \geq 1$ . I thought about using LTE 2 times for the numerator but I got stuck.
math.enthusiast9
  • 1,031
  • 3
  • 14
2
votes
4 answers

Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$

I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc. So that means that $\frac{10^a-1}{9}$ must be divisible by 3. $\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1…
O.S.
  • 602
0
votes
0 answers

Stabilization of one new trailing digit of $a^{10^b}$ as $b$ increases, for every $a$ not divisible by $10$

Let $a,b \in \mathbb{N}-\{0,1\}$ be such that $10 \nmid a$ . Then, define $S_a(b) := \nu_{10}(a^{10^{b+1}} - a^{10^b})$ and $D_a(b) := S_a(b) - S_a(b-1)$ . So, $S_a(b)$ counts the number of common trailing digits between $a^{10^{b+1}}$ and…
Marco Ripà
  • 1,469
-1
votes
1 answer

To prove If $a ≡ b$ mod $m$, then $a^2 ≡ b^2$ mod $m^2$

I am tying to disprove it but I feel my proof is lacking, please let me know what I can improve on from here, thanks!: If $a ≡ b$ mod $m$ , then $m|a-b$ i.e. $a-b=km$ , $k$ is an int. Square both sides we get: $(a-b)^2=k^2m^2\implies…
quadingle the 2nd
  • 19