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

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67
votes
5answers
10k views

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

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
16
votes
3answers
827 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 = ...
26
votes
9answers
6k 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 $a^n$ ...
14
votes
7answers
12k 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 ...
7
votes
2answers
910 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 ...
6
votes
3answers
9k 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
5answers
441 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 $d=\gcd(a,b)$? Note $\ $ Some of the answers below were merged from this ...
4
votes
3answers
906 views

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

I need to prove that $z\gcd(a,b)=\gcd(za,zb)$. 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 please give me ...
9
votes
2answers
3k views

Derive a formula to find the number of trailing zeroes in $n!$ [duplicate]

Possible Duplicate: How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? I know that I have to find the number of factors of $5$'s, $25$'s, ...
15
votes
17answers
13k 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 ...
9
votes
5answers
568 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 ...
2
votes
2answers
147 views

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some ...
58
votes
7answers
8k views

What makes $9$ special?

I don't know if this is a well know fact but I have observed that every number no matter how large that is equally divided by $9$, will equal $9$ if you add all the numbers it is made from until there ...
52
votes
16answers
16k views

For any prime $p > 3$, why is $p^2-1$ always divisible by 24?

I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.
24
votes
11answers
8k views

Division of Factorials

I have a partition of a positive integer $(p)$. How can I prove that the factorial of $p$ can always be divided by the product of the factorials of the parts? As a quick example $\frac{9!}{(2!3!4!)} ...
10
votes
4answers
527 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 $a,b,c,z \in \mathbb{Z}$).
2
votes
3answers
2k views

GCD Proof with Multiplication: gcd(ax,bx) = x$\cdot$gcd(a,b)

I was curious as to another method of proof for this: Given $a$, $b$, and $x$ are all natural numbers, $\gcd(ax,bx) = x \cdot \gcd(a,b)$ I'm confident I've found the method using a generic common ...
43
votes
3answers
2k views

How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? [duplicate]

Possible Duplicate: Highest power of a prime $p$ dividing $N!$ How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
4
votes
2answers
429 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\\ ...
11
votes
2answers
3k 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 ^ ...
5
votes
1answer
205 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 .
23
votes
6answers
18k 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 ...
10
votes
3answers
212 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 ...
2
votes
2answers
8k 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 ...
3
votes
2answers
2k views

If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$.

Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$ I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$ I thought setting $x = (a+b)$ and $y = ...
2
votes
4answers
621 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 ...
8
votes
7answers
14k 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.
1
vote
6answers
5k views

suppose $\gcd(a,b)= 1$ and $a$ divides $bc$. Show that $a$ must divide $c$.

Well I thought this is obvious. since $\gcd(a,b)=1$, then we have that $a$ does not divide $b$ AND $a$ divides $bc$. this implies that $a$ divides $c$. done. but apparently this is wrong. help ...
12
votes
5answers
4k 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. ...
8
votes
6answers
7k 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) ...
10
votes
3answers
509 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, $a$ ...
5
votes
6answers
2k views

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

I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. ...
6
votes
3answers
868 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 ...
5
votes
5answers
5k 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 ...
9
votes
1answer
997 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 ...
12
votes
6answers
6k views

How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$?

I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
1
vote
1answer
275 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 ...
13
votes
7answers
747 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.
1
vote
5answers
88 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.
6
votes
3answers
353 views

For integers $a$ and $b$, $ab=\text{lcm}(a,b)\cdot\text{hcf}(a,b)$

I was reading a text book and came across the following: Important Results (This comes immediately after LCM:) If 2 [integers] $a$ and $b$ are given, and their $LCM$ and $HCF$ are $L$ and ...
45
votes
8answers
5k views

What is $\gcd(0,0)$?

What is the greatest common divisor of $0$ and $0$? On the one hand, Wolfram Alpha says that it is $0$; on the other hand, it also claims that $100$ divides $0$, so $100$ should be a greater common ...
25
votes
5answers
883 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 ...
7
votes
3answers
761 views

Proof of Wolstenholme's theorem

According to the theorem, if $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$ then we have to prove that $r\equiv0 \pmod{p^2}$. (Given $p>3$, otherwise ...
10
votes
5answers
356 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 ...
6
votes
1answer
367 views

Showing $\gcd(2^m-1,2^n+1)=1$

A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement If $m$ is odd then ...
6
votes
8answers
396 views

If a prime $p\mid ab$, then $p\mid a$ or $p\mid b$

If a prime number $p$ is a divisor of a product $ab$, $p$ has to be a divisor of $b$ or $a$. How can I demonstrate this theorem? I demonstrated this theorem on one way using Bezout's theorem in an ...
5
votes
3answers
5k 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 $n = xy$. $2^{xy} - 1 = (2^x)^y - 1$ $ = ...
1
vote
7answers
915 views

Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers.

Let $m$ and $n$ be two integers. Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers. I think I have to use the contrapositive to solve this. So I assume $\neg ...
6
votes
3answers
2k 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 ...
3
votes
4answers
648 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 ...