This tag is for questions about divisibility, that is, determining when one thing is a multiple of another thing.

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40
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3answers
7k views

Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
3
votes
2answers
325 views

Divisibility for 7

I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let $n = ...
12
votes
7answers
7k views

The product of n consecutive integers is divisible by n factorial

How can we prove that the product of n consecutive integers is divisible by n factorial? Note: In this subsequent question and the comments here the OP has clarified that he seeks a proof that ...
22
votes
5answers
4k views

Why $a^n - b^n$ is divisible by $a-b$?

I did some mathematical induction problems on divisibility $9^n$ $-$ $2^n$ is divisible by 7. $4^n$ $-$ $1$ is divisible by 3. $9^n$ $-$ $4^n$ is divisible by 5. Can these be generalized as $a^n$ ...
5
votes
2answers
679 views

Divisibility criteria for $7,11,13,17,19$

A number is divisible by $2$ if it ends in $0,2,4,6,8$. It is divisible by $3$ if sum of ciphers is divisible by $3$. It is divisible by $5$ if it ends $0$ or $5$. These are simple criteria for ...
5
votes
3answers
6k views

The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

How can we prove, without using the properties of binomial coefficients, the product of n consecutive integers is divisible by n factorial?
10
votes
3answers
267 views

Show that $\gcd\left(\frac{a^n-b^n}{a-b},a-b\right)=\gcd(n d^{n-1},a-b)$

How to show that $$ \gcd\bigg( {a^n-b^n \over a-b} ,a-b\bigg )=\gcd(n d^{n-1},a-b ) $$ $a,b\in \mathbb Z$ where $d=\gcd(a,b)$?
2
votes
16answers
3k views

How to Prove the divisibility rule for $3$

The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three. This is quite helpful for determining if really large ...
11
votes
4answers
372 views

Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$

I can't crack this one. Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ (the only constraint is that $a,b,c,z \in \mathbb{Z}$).
4
votes
3answers
583 views

How to prove that $z\cdot\text{gcd}(a,b)=\text{gcd}(za,zb)$

I need to prove that $z \cdot \text{gcd}(a,b)=\text{gcd}(za,zb)$. I tried a lot, for example, looking at set of common divisors of the two sides, but I can't conclude anything from that. Can you ...
10
votes
2answers
2k views

$\gcd(b^x - 1, b^y - 1, b^ z- 1,…) = b^{\gcd(x, y, z,…)} -1$ [duplicate]

Possible Duplicate: Number theory proving question? Dear friends, Since $b$, $x$, $y$, $z$, $\ldots$ are integers greater than 1, how can we prove that $$ \gcd (b ^ x - 1, b ^ y - 1, b ^ ...
11
votes
6answers
3k views

Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$.

Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. ...
5
votes
5answers
279 views

Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$

Somewhere on Stack Exchange I saw the equation $$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$ I had never seen this before, so I started trying to prove it. Without success... Can anyone explain me (so ...
6
votes
3answers
690 views

Why (directly!) does every number divide 9, 99, 999, … or 10, 100, 1000, …, or their product?

A curiosity that's been bugging me. More precisely: Given any integers $b\geq 1$ and $n\geq 2$, there exist integers $0\leq k, l\leq b-1$ such that $b$ divides $n^l(n^k - 1)$ exactly. The ...
23
votes
6answers
12k views

If $a^2$ divides $b^2$, then $a$ divides $b$

Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$. Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out ...
9
votes
1answer
759 views

If $\gcd(a,b)=d$, then $\gcd(ac,bc)=cd$?

$A$ an integral domain, $a,b,c\in A$. If $d$ is a greatest common divisor of $a$ and $b$, is it true that $cd$ is a greatest common divisor of $ca$ and $cb$? I know it is true if $A$ is a UFD, but ...
10
votes
3answers
176 views

Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$

Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$. I am having trouble showing this. I've seen a similar ...
5
votes
1answer
108 views

$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$

Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$ Thanks in advance .
1
vote
4answers
272 views

Prove that $(ma, mb) = |m|(a, b)$

I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max ...
18
votes
3answers
609 views

Prove $n\mid \phi(2^n-1)$

If $2^p-1$ is a prime,(thus $p$ is a prime,too) then $p\mid 2^p-2=\phi(2^p-1).$ But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is.Such as $4\mid \phi(2^4-1)=8.$ If we denote ...
3
votes
2answers
160 views

Extended Euclidean Algorithm problem

I'm confused about how to do the extended algorithm. For example, here's the gcd part gcd(8000,7001) $$\begin{align}8000 &= 7001\cdot1 + 999\\ 7001&=999\cdot 7+8\\ 999&=8\cdot 124+7\\ ...
12
votes
7answers
527 views

How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$

If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can´t find a way to use any of the elemental divisibility and gcd theorems to find them.
6
votes
5answers
4k views

Prove $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ if $\gcd(a,b) = 1$

I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$ $\gcd(a+b, a-b) > = 1$ or $\gcd(a+b, a-b) = 2$ holds. I think the first step should look something like this: $d = \gcd(a+b, a-b) ...
9
votes
3answers
450 views

If $R$ is a commutative ring with identity, and $a, b\in R$ are divisible by each other, is it true that they must be associates?

Thank you very much! My problem is: If $R$ is a commutative ring with identity, and $a, b$ are its elements that are divisible by each other, is it true that they must be associates? Here, $a$ ...
1
vote
1answer
216 views

How to prove Greatest Common Divisor using Bézout's Lemma

The problem is to prove the following If $\gcd(a,b) = c$, then $\gcd(a^m, b^m) = c^m$ I know that this can be solved easily by proving that $c\mid a \implies c^m \mid a^m$ and $c\mid b \implies ...
6
votes
7answers
6k views

If gcd (a,b)=1 and gcd (a,c)=1, then gcd (a,bc)=1

How do I go about proving this? If gcd (a,b)=1 and gcd (a,c)=1, then gcd (a,bc)=1. I'm very confused with gcd proofs.
4
votes
5answers
4k views

Simple Proof by induction: “9 divides $n^3 + (n+1)^3 + (n+2)^3$”

I'm trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have: Base case: P(1) = (1) + (8) + (27) = 36, 36 can be divided by 9 ...
2
votes
2answers
112 views

Divisibility of prime numbers

I have this exercise in my worksheet in the discrete mathematics course.I don't understand the part that deals with prime numbers in integer-divisibility. "Show that for a prime number $p$, if a ...
1
vote
4answers
444 views

Divisibility and the Fibonacci sequence

While studying the Fibonacci sequence I encountered this problem in the handout, and I can not understand how to do it. Show that if the Fibonacci sequence has a term divisible by a natural number ...
1
vote
5answers
78 views

Prove if $a\mid b$ and $b\mid a$, then $|a|=|b|$ , $a, b$ are integers.

Form the assumption, we can say $b=ak$ ,$k$ integer, $a=bm$, $m$ integer. Intuitively, this conjecture makes sense. But I can't make further step.
1
vote
2answers
4k views

If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$

Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$. This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was ...
14
votes
7answers
698 views

$n^5-n$ is divisible by $10$?

I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that ...
1
vote
3answers
66 views

If $\gcd (a,b) = 1$, what can be said about $\gcd (a+b,a-b)$?

Suppose $a, b \in \mathbb{Z}$, $a > b$, and $\gcd (a,b) = 1$. What can be said about $\gcd (a+b,a-b)$? Is it true in general that $\gcd (a+b,a-b) \leq 2$?
4
votes
3answers
1k views

Show that $\gcd(a,bc)=1$ if and only if $\gcd(a,b)=1$ and $\gcd(a,c)=1$

Show that $\gcd(a,bc)=1$ if and only if $\gcd(a,b)=1$ and $\gcd(a,c)=1$. I am new at proofs and I think I should use Euclid's Lemma which states "If $p$ is a prime that divides $ab$, then $p$ divides ...
2
votes
3answers
64 views

Proof that if $\gcd(a,b) = 1$ and $a\mid n$ and $b\mid n$, $ab \mid n$

I'm learning about properties of greatest common divisors, specifically when two numbers are relatively prime. The exercise I'm working through is : Suppose that $\gcd(a,b) = 1$ and that $a\mid ...
0
votes
1answer
59 views

Is it possible to find $n-1$ consecutive composite integers

Given an integer $n\geq 2$ ,can we always find an integer $m$ such that each of the $n-1$ consecutive integers $m+2,m+3,.....,m+n$ are composite?
11
votes
2answers
477 views

How does one attack a divisibility problem like $(a+b)^2 \mid (2a^3+6a^2b+1)$?

In my current line of investigation, I am running into [many] divisibility questions like the one in the title, i.e. $$ (a+b)^2 \mid (2a^3+6a^2b+1), \qquad(\star) $$ where $a > b \ge 1$ are ...
9
votes
5answers
325 views

Prove that $b\mid a \implies (n^b-1)\mid (n^a-1)$

Given natural numbers $a,b,n$, prove $b\mid a \implies (n^b-1)\mid (n^a-1)$. I tried the simple method of beginning with $b\mid a \implies \exists k \in \mathbb{N} $ such that $bk=a$ and then ...
7
votes
3answers
1k views

Divisibility rules and congruences

Sorry if the question is old but I wasn't able to figure out the answer yet. I know that there are a lot of divisibility rules, ie: sum of digits, alternate plus and minus digits, etc... but how can ...
16
votes
2answers
839 views

Proving there are an infinite number of pairs of positive integers $(m,n)$ such that $\frac{m+1}{n}+\frac{n+1}{m}$ is a positive integer

The question is: Show that there are an infinite number of pairs $(m,n): m, n \in \mathbb{Z}^{+}$, such that: $$\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{Z}^{+}$$ I started off approaching this ...
8
votes
3answers
481 views

Proof of Wolstenholme's theorem.?

According to the theorem : $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{p-1} =\frac{r}{q}$$ And we have to prove that $r= 0 \pmod{p^2}$. (Given $ p>3$, ...
4
votes
1answer
269 views

Divisibility of $2^n - 1$ by $2^{m+n} - 3^m$.

For what values of $m,n$ natural, do $2^n - 1$ is divisible by $2^{m+n} - 3^m$? Thank you very much.
2
votes
4answers
174 views

Division of $q^n-1$ by $q^m-1$, in Wedderburn's theorem

I need this for a proof of Wedderburn's theorem: $$q^m - 1 | q^n - 1 \quad \Rightarrow \quad m|n$$ with $q>1 \in \mathbf{N}$ and $m,n \in \mathbf{N}$. I'd also like to know if it works the other ...
9
votes
3answers
384 views

Does $a^n \mid b^n$ imply $a\mid b$?

Does $a^n \mid b^n$ imply $a\mid b$? I think it does but haven't been able to prove it. I don't know much number theory so an elementary answer would be great.
2
votes
3answers
163 views

Prove that distinct Fermat Numbers are relatively prime [duplicate]

The Fermat numbers are defined by $F_m = 2^{2^m} + 1$. Prove that for $m \ne n$ we have $(F_m, F_n) = 1$. I have to first prove that $F_{m+1} = F_0F_1 \cdots F_m + 2$ by representing $F_{m+1}$ in ...
2
votes
3answers
113 views

Divisor in $\mathbb{C}[X]$ $\implies$ divisor in $\mathbb{R}[X]$?

let $P \in \mathbb{R}[X]$ be a real polynomial divisible by a polynomial $Q \in \mathbb{R}[X]$ in $\mathbb{C}[X]$. How can I easily show that $P$ is also divisible by $Q$ in $\mathbb{R}[X]$? A simple ...
5
votes
3answers
2k views

If $2^n - 1$ is prime from some integer $n$, prove that n must also be prime.

I understand the idea of the proof. I just want to make sure I wrote my proof well. Suppose $n$ is not prime. Then $\exists x,y \in \mathbb{Z}$ such that $n = xy$. $2^{xy} - 1 = (2^x)^y - 1$ $ = ...
0
votes
3answers
950 views

How to show that $\gcd(n! + 1, (n + 1)! + 1) \mid n$?

Let $n$ be a positive integer, $n!$ denotes the factorial of $n$. Let $d = \gcd(n! + 1, (n + 1)! + 1)$. Show that $d$ divides $n$. (Hint: notice that $(n+1)(n!+1) = (n+1)!+n+1$)
6
votes
3answers
1k views

Prove that every positive integer $n$ is a unique product of a square and a squarefree number

I am trying to prove that for every integer $n \ge 1$, there exists uniquely determined $a > 0$ and $b > 0$ such that $n = a^2 b$, where $b$ is squarefree. I am trying to prove this using the ...
4
votes
3answers
403 views

Prove that $(a+1)(a+2)…(a+b)$ is divisible by $b!$ [duplicate]

The problem is following, prove that: $$(a+1)(a+2)...(a+b)\text{ is divisible by } b!\text{ for every positive integer a,b}$$ I've tried solving this problem using mathematical induction, but I ...