# Tagged Questions

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

4answers
1k views

### Proving that $\gcd(a,b) = as + bt$, i.e., $\gcd$ is a linear combination. [duplicate]

For any nonzero integers $a$ and $b$, there exist integers $s$ and $t$ such that $\gcd(a,b) = as + bt$. Moreover, $\gcd(a,b)$ is the smallest positive integer of the form $as + bt$. I ...
2answers
91 views

### cancelling out before evaluation of variable

I'm been working on a theory, though my math is weak. Let's say I've managed to determine that I can arrive at an answer A by always using the formula BCD / D. Of ...
2answers
259 views

1answer
4k views

### Sum of GCD(k,n)

I want to find this $$\sum_{k=1}^n \gcd(k,n)$$ but I don't know how to solve. Does anybody can help me to finding this problem. Thanks.
2answers
140 views

### $1^k+2^k+3^k+…+(p-1)^k$ always a multiple of $p$?

I would appreciate if somebody could help me with the following problem: Q: For any prime number $p(p\geq 3), k=1,2,3,...,p-2$, why is $$1^k+2^k+3^k+...+(p-1)^k$$ always a multiple of $p$ ?
1answer
73 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?
3answers
82 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 $\gcd (a+b,a-b) \leq 2$?
3answers
2k 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 ...
3answers
548 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 ...
6answers
324 views

### Prove that $6$ divides $n(n + 1)(n + 2)$

I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows: Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$. ...
4answers
655 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 ...
3answers
90 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 $a\mid n$...
4answers
2k views

1answer
109 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 ...
3answers
128 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 ...
3answers
606 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 ...
4answers
216 views

### If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$.

If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. Also... $a,b$ and $n$ are natural numbers. I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from ...
4answers
3k views

### Prove $\gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $\gcd(a,b) = 1$

Assuming that $\gcd(a,b) = 1$, prove that $\gcd(a+b,a^2+b^2) = 1$ or $2$. I tried this problem and ended up with $$d\mid 2a^2,\quad d\mid 2b^2$$ where $d = \gcd(a+b,a^2+b^2)$, but then I am stuck; ...
4answers
4k views

3answers
135 views

### Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ so on , then $a=b$?

Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ that is $a^{2n-1}\mid b^{2n} ; b^{2n}\mid a^{2n+1} , \forall n \in \mathbb Z^+$ , then is it true that $a=b$ ?
2answers
766 views

### Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n [duplicate]

Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n Hi everyone, for the proof to the above question, Can I assume that since $(a, b) = 1$,...
3answers
1k 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 $(n+1)(n!+1) = (n+1)!+n+1$)
2answers
143 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&=& p_1^{\...
5answers
1k views

### Prove 24 divides $u^3-u$ for all odd natural numbers $u$

At our college, a professor told us to prove by a semi-formal demonstration (without complete induction): For every odd natural: $24\mid(u^3-u)$ He said that that example was taken from a high ...
5answers
998 views

### Prove by induction that $2^{2n} – 1$ is divisible by $3$ whenever n is a positive integer.

I am confused as to how to solve this question. For the Base case $n=1$, $(2^{2(1)} - 1)\,/\, 3 = 1$, base case holds My induction hypothesis is: Assume $2^{2k} -1$ is divisible by $3$ when $k$ is a ...
1answer
9k views

### Prove by mathematical induction that $n^3 - n$ is divisible by $3$ for all natural number $n$

I'm working on a task where I'm a bit unsure if the answer I've got is correct. Here is the task: Show by induction that the following assertion is true for all natural numbers $n$ \$n^3 - ...