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

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2
votes
1answer
19 views

Prove that if $a, b, n\in \mathbb{N}, n\geq2\longrightarrow \sqrt[\leftroot{-2}\uproot{2}n]{a}\in \mathbb{Q} \iff a=b^n$.

Prove that if $a, b, n\in \mathbb{N}, n\geq2\longrightarrow \sqrt[\leftroot{-2}\uproot{2}n]{a}\in \mathbb{Q} \iff a=b^n$. I'm at a complete loss here, I tried using the order of a prime function but ...
1
vote
0answers
61 views

Fermat's Little Theorem and prime divisors

Let $a,b\in\Bbb N$ and $a+b$ be an even number. Assume $a^2 - b^2 - a$ is an exact square, say $c^2$. Let $m = \frac {a+b}2$ and $n = \frac {a-b}2$. Then, $$(4m-1)(4n-1) = 4(4mn-m-n) + 1 = ...
2
votes
0answers
15 views

Example of a domain where all irreducibles are primes and that is not a GCD domain

One has the following relations for a domain $R$: $R$ GCD domain $\Rightarrow$ All irreducible elements are prime $R$ PID $\Rightarrow$ $(R$ GCD domain $\land$ $R$ statisfies ACCP$)$ $R$ UFD ...
2
votes
1answer
34 views

An expression with gcd and abs is transformed magically!

There's a problem to calculate $\sum^{n}_{i=1}\sum^{m}_{j=1}\frac{|i-j|}{\gcd(i,j)}$, whose tutorial gives the following transformation I really don't understand. ...
5
votes
1answer
48 views

Pythagorean Triples : Is every positive integer $\gt$ $2$ part of at least one Pythagorean triple?

I was doing some basic number theory problems from Rosen and came across this problem: Show that every positive integer $\gt$ $2$ is part of at least one ...
5
votes
1answer
85 views

Pythagorean Triples : Show that exactly one of $x$, $y$, and $z$ is divisible by $5$

I was doing some basic number theory problems from Rosen and came across this problem: Show that if $(x, y,z)$ is a primitive Pythagorean triple, then exactly one of $x$, $y$, and $z$ is divisible ...
10
votes
5answers
292 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 ...
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 ...
2
votes
1answer
53 views

Number of pairs $(A,B)$ with $\gcd(A,B)=B, A \ne B^2$ with $A,B \le n$

How many pairs $(A,B)$ of integers up to $n$ are there such that $\gcd(A,B)=B$, not counting those pairs where $B^2=A$? If we consider $n = 5$ we have $25$ possible pairs. They are ...
-3
votes
2answers
33 views

List the elements of a set [on hold]

Consider the universal set $N$, $$A = \{m: m\ |\ 16\}$$ and $$B = \{n: n \le 16 \text{ and } n \equiv 17 \mod 3\}.$$ List the elements of A, list the elements of B.
1
vote
0answers
26 views

$\max{a_1}$ in $(x_n)_{n\in\mathbb{N}}$

Let be $x_1,x_2,x_3,\ldots,$ a sequence of positive integers. Suposse the folowing conditions are true for all $n\in\mathbb{N}$ $n|x_n$ $|x_n-x_{n+1}|\leq 4$ Find the maximun value of $x_1$ I ...
0
votes
5answers
72 views

Mathematical induction [duplicate]

Prove that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ where $n$ is a nonnegative integer. I have seen many questions on this site that contain the answer to this problem and I already know the solution, ...
1
vote
5answers
96 views

Proving $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ [duplicate]

I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly. ...
0
votes
3answers
405 views

Mathematical induction prove that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ . [duplicate]

How can I use mathematical induction to prove that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a nonnegative integer?
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 ...
0
votes
1answer
33 views

Find two numbers, given their greatest common divisor and least common multiple [on hold]

Highest common factor (HCF) of two numbers is $20$. Least common multiple (LCM) of the same two numbers is $420$. Both numbers are higher than $50$. Find the $2$ numbers. I used factorising trees ...
-1
votes
0answers
24 views
5
votes
4answers
1k views

How many integers in the range [1,999] are divisible by exactly 1 of 7 and 11?

This is a question in Kenneth Rosen's Discrete Mathematics textbook 6th edition. I haven't had trouble with any other counting problems regarding "how many numbers in range [x,y] have divisibility ...
2
votes
1answer
37 views

Proving $k$ is divisible by $3$ iff the sum of the digits of $k$ is divisible by 3 [duplicate]

I am trying to prove that $k$ is divisible by $3$ iff the sum of the digits of $k$ is divisible by 3 for all $k \in Z$. I am not even sure what tags to use because I am not sure of right methods to ...
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 ...
3
votes
1answer
37 views

Decide if there exist $a$ and $b \in \mathbb{Z}$ such that $a^2=2b^2$.

Decide if there exist $a$ and $b \in \mathbb{Z}$ such that $a^2=2b^2$. $a,b \neq 0$ We have to solve this kinds of problems using the order of a prime function: $v_p(a) \in \mathbb{Z}$ which tells ...
6
votes
2answers
87 views

Prove or disprove : $a^3\mid b^2 \Rightarrow a\mid b$

I think it's true, because I can't see counterexamples. Here's a proof that I am not sure of: Let $p_1,p_2,\ldots, p_n$ be the prime factors of $a$ or $b$ \begin{eqnarray} a&=& ...
2
votes
3answers
72 views

$5 \nmid 2^{n}-1$ when $n$ is odd

I want to prove that $$5 \nmid 2^{n}-1$$ where $n$ is odd. I used Fermat's little theorem, which says $2^4 \equiv 1 \pmod 5$, because $n$ is odd then $4 \nmid n$ , so it is done. can you check it ...
0
votes
3answers
40 views

Find $GCD(n^2+1,n+1)$

$GCD(n^2+1,n+1)$, $n\in \mathbb{N}$ What I did: $n^2+1=(n-1)(n+1) + 0$ So I thought $(n^2+1:n+1)=n+1$ But that doesn't seem to be the case: $n=2$ $n^2+1=5$ $n+1=3$ $GCD(5,3)=1$ Why is the ...
1
vote
2answers
12 views

Prove the order of the group homomorphism of an element divides the order of the element.

Let $\phi : G \rightarrow H$ be a group homomorphism. Prove $\forall g \in G$, the order of $\phi(g)$ divides $g$. I've gotten to the point where I've shown that if, $ord(\phi(g)) < ord(g)$ then ...
6
votes
2answers
723 views

Fibonacci and Lucas identity

By the trial and error method I have observed the following identity by taking some numerical values. Those are $F_m$|$L_n$ is valid only if one of the following holds. a) $m = 1$ or $m =2$ b) $m ...
1
vote
1answer
357 views

Proof by induction involving fibonacci numbers

The Fibonacci numbers are defined as follows: $f_0=0$, $f_1=1$, and $f_n=f_{n−1}+f_{n−2}$ for $n≥2$. Prove each of the following three claims: (i) For each $n≥0$, $f_{3n}$ is even. (ii) For each ...
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 $a,b,c,z \in \mathbb{Z}$).
3
votes
3answers
57 views

Let $p$ be a prime. Why is ${p^mn \choose p^m}$, where $p \nmid n$, not divisible by $p$? [duplicate]

Let $p$ be a prime. Why is ${p^mn \choose p^m}$, where $p \nmid n$, not divisible by $p$? $${p^mn \choose p^m} = \frac{(p^mn)!}{p^m!(p^mn-p^m)!} = ...
1
vote
2answers
28 views

prove that $\forall k\in \mathbb Z$ if $(m,n)=1$(coprimes) then $(k,m,n)=1$

Prove that $\forall k\in \mathbb Z$ if $(m,n)=1$(coprimes) then $(k,m,n)=1$ I did this by contradiction: let $d=(k,m,n)$ so that $d\neq 1$; by definition of greatest common divisor $d|k, d|m, d|n$ ...
1
vote
2answers
48 views

For a primitive Pythagorean triple $(a, b, c)$, is it always true that $\gcd(a,b) = \gcd(b,c) = \gcd(a,c) = 1$?

Let $(a, b, c)$ be a primitive Pythagorean triple. I know that $\gcd(a,b,c) = 1$. Is it always true that $\gcd(a,b) = \gcd(b,c) = \gcd(a,c) = 1$?
1
vote
6answers
143 views

Euclidean Algorithm Question

So I have been asked to find $d=(a,b)$ when $a=1109$ and $b=4999$ and express $d$ as a linear combination of $a$ and $b$ Well I have worked out that $d=1$ but I am struggling to express $d$ as a ...
1
vote
4answers
78 views

Prove that if $3|mn$, then $3|m$ or $3|n$

I am trying to prove this for integers $m$ and $n$. I tried to reach prove that $3|m$ by assuming that 3 does not divide $n$, but this is such a basic assumption of mine already that it is hard for ...
0
votes
3answers
27 views

Binary remainder not equal to the decimal remainder

I am having a weird result. I am dividing the binary number $10101010100000$ by $10011$. In binary division. I get $R= 0100$ which is 4. However, If I consider the decimal representation of the ...
2
votes
1answer
38 views

How to show that $\mathbb Z+x \mathbb Q[x]$ is a GCD domain ?

How to show that $\mathbb Z+x \mathbb Q[x]$ is a Bezout domain , that is sum of two principal ideals is again a principal ideal ? Or at least , how to show that it is a GCD domain ? ( This will then ...
5
votes
7answers
291 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 ...
-3
votes
1answer
40 views

can a positive integer have 2010 facters? [closed]

For any positive integer 𝑛, let 𝑓(𝑛) be the number of distinct positive integer factors of 𝑛 including 1 and n. Which of the following is (are) true? I There is a positive integer 𝑛0 such that ...
1
vote
2answers
89 views

how $1/0.5$ is equal to $2$?

My question is how $1/0.5$ is equal to $2$. I am not asking the mathematical justification that $1/0.5=10/5=2$. I know all this. I just want to know how it is two... a lay man justification. ...
1
vote
1answer
33 views

Looking for an example of a GCD domain which is not UFD

I know that every UFD (unique factorization domain ) is a GCD domain i.e. g.c.d. of any two elements , not both zero , exists in the domain . I am looking for an example of a GCD domain which is not ...
1
vote
1answer
86 views

Prove that we always have $ 2n \mid \varphi(m^n+p^n) $

For each $ a ∈ \Bbb N^*$, denoted by $\varphi (a) $ is the number of positive integers not exceeding $a$ and coprime to $a$. Let $n, m, p ∈ \Bbb N^*, m \ne p$. Prove that we always have $2n \mid ...
2
votes
0answers
28 views

Polynomials and Divisibility Rule.

The question is this - If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $h(x)=xf(x^3)+x^2g(x^6)$ is divisible by $x^2+x+1$, then which of the following are true? 1. $f(1)=g(1)$ ...
1
vote
1answer
30 views

Finding a natural number $k>1$ such that $k$ divides $(26+35n)$ and $(3+7n)$

I am trying to find a natural number $k>1$ such that $k$ divides $(26+35n)$ and $k$ divides $(3+7n)$ for some integer $n$. I know that $(ka)=(26+35n)$ for some $a \in Z$ and $(kb)=(3+7n)$ for some ...
2
votes
0answers
21 views

Show that $\gcd(x_1,…,x_k,x_{k+1})=\gcd(\gcd(x_1,…,x_k),x_{k+1})$

I would really appreciate if you could check my proof. Thank you. Show that $\gcd(x_1,...,x_k,x_{k+1})=\gcd(\gcd(x_1,...,x_k),x_{k+1})$ Let $\gcd(x_1,...,x_k,x_{k+1})=a$. We first show that $a$ ...
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 = ...
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 ...
40
votes
3answers
7k views

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

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
0
votes
1answer
42 views

If $m\mid n$ then $p^m-1\mid p^n-1$ [duplicate]

I know $m$ ,$n$ are two positive integer numbers such that $m\mid n$. If $p$ is a prime number, I want to show $p^m-1\mid p^n-1$.
0
votes
2answers
42 views

Find $a,b,c$ if $\gcd(a,b,c)=10$ and $\text{lcm}(a,b,c)=100$ [closed]

Find all natural numbers $a,b,c$, so that $\gcd(a,b,c)=10$ and $\text{lcm}(a,b,c)=100$. Any ideas how to approach this problem?