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

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32
votes
3answers
6k views

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

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
22
votes
5answers
3k 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 $a^n$ ...
2
votes
2answers
145 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 = ...
5
votes
2answers
614 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 ...
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. ...
6
votes
3answers
647 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 ...
21
votes
5answers
11k 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 ...
8
votes
1answer
691 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 ...
3
votes
3answers
478 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 ...
1
vote
4answers
190 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 ...
10
votes
4answers
324 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 $a,b,c,z \in \mathbb{Z}$).
6
votes
5answers
3k 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: $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$ ...
12
votes
7answers
471 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
1answer
188 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 ...
5
votes
7answers
5k 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
4answers
380 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 ...
14
votes
3answers
507 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 ...
6
votes
1answer
96 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 .
14
votes
7answers
670 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 ...
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 ...
0
votes
1answer
83 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 ...
10
votes
2answers
460 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 ...
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 ...
9
votes
5answers
312 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 ...
8
votes
3answers
428 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
259 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 ...
15
votes
2answers
791 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
433 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
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 ...
0
votes
3answers
865 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
957 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
368 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 ...
1
vote
5answers
254 views

$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement.

"Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$" Resolution: $$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid ...
0
votes
5answers
187 views

Proving that if $\;5\mid(a+11)\;$ and $\;5\mid(16-b),\;$ then $\;5\mid(a+b)$

Can you please help me a bit with this question? How do we show that $\;$ if $\;\;5\mid(a+11)\;$ and $\;5\mid(16-b),\;\;$ then $\;5\mid(a+b)\;$? Thanks a lot!
13
votes
2answers
283 views

Prove $6 \nmid [\left( \sqrt[3]{28} - 3 \right)^{-n}]$

Prove that: $$6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.$$ ($\lfloor x\rfloor$ = largest integer not exceeding $x$) I am very bad as English and ...
11
votes
2answers
601 views

Does $n \mid 2^{2^n+1}+1$ imply $n \mid 2^{2^{2^n+1}+1}+1$?

There are two ways to try to prove this. One is in the title, the other is its de Morgan counterpart: $n \nmid 2^{2^{2^n+1}+1}+1 \implies n \nmid 2^{2^n+1}+1$. Disproving it requires only one example ...
0
votes
2answers
4k views

How many positive integers less than 1000 are divisible [closed]

How many positive integers less than 1000 c) are divisible by both 7 and 11? d) are divisible by either 7 or 11? e) are divisible by exactly one of 7 and 11?
11
votes
2answers
653 views

What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?

What is the remainder when $$1! + 2! + 3! +\cdots+ 1000!$$ is divided by $12$. I have tried to find the answer using the Binomial Theorem but that doesn't help. How will we do this? Please help.
9
votes
3answers
366 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.
3
votes
1answer
61 views

Determine the divisibility of a given number without performing full division

My question is slightly more complicated than what's implied on the title, so I will start with an example. Given any number $N$ on base $10$, we can easily determine whether or not $N$ is divisible ...
3
votes
3answers
464 views

Number of divisors

How can I find number of divisors of N which are not divisible by K. ($2 \leq N$, $k \leq 10^{15})$ One of the most easiest approach which I have thought is to first calculate total number of ...
1
vote
5answers
69 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.
9
votes
6answers
258 views

Understanding the proof of a formula for $p^e\Vert n!$

This is a proof from a book on number theory I'm reading. I'm having a hard time following. I think there's a variable here that means two different things at two different times... Theorem: If n is ...
6
votes
3answers
12k views

Find the largest prime factor

I just "solved" the third Project Euler problem: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? With this on Mathematica: ...
4
votes
5answers
162 views

If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$ [duplicate]

Prove or disprove 'If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$, with $a,b\not= 0$' I need to prove this statement. I think it is true and also the converse is true. I took some examples such as ...
4
votes
3answers
1k 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 $n = xy$. $2^{xy} - 1 = (2^x)^y - 1$ $ = ...
3
votes
5answers
119 views

How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$?

I've been asked to show that: $n(n+1)(2n+1) \equiv 0 \pmod 6$ I found in a previous question that: $n(n+1)$ was divisible by $2$ and resulted in an even number e.g $n(n+1) \equiv 0 \pmod 2$ so I ...
3
votes
2answers
3k views

Proof of $\gcd(a,b)=ax+by$

Here is my proof of $\gcd(a,b)=ax+by$ for $a, b, x, y \in \mathbb{Z}$. Am I doing something wrong? Are there easier proofs? $a,b \in \mathbb{Z}, g=\gcd(a,b)$ and suppose $g \neq ax + by$. Let $c$ be ...