Question is:     If $a$ and $b$ are prime to each other and $n$ is prime then prove that $\frac{a^n+b^n}{a+b}$ and   $a+b$ have no common factor unless $a+b$ is a multiple of $n$.   This is what I thought so far:  $a^n$ is also prime to $b^n$ because of the theorem (If $a$ is prime to $b$ then $a^n$ is also prime to $b^n$), Also $a^n$ and $b^n$ is a divisor of $(a+b)$ therefore by theorem (If $a$ is prime to $b$, and each of these numbers is a divisors of $N$, then $ab$ is a divisor of $N$) $a^nb^n$ is a divisor of $(a+b)$.  Thereafter no idea, How should I proceed further if I'm thinking in a right way and If not, How should I prove this?  Source: Higher Algebra by Barnard and Child
24558
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 . The answers (and their comments) may depend on context provided in that question.
92631
The question is:      Prove that if $p \in \mathbb{Z}$ is irreducible, then $p$ is also prime.    Irreducible is defined as follows: $n \in \mathbb{Z} $ is irreducible iff the only way to write $n=ab$ for some integers $a$ and $b$ is if $a = \pm1$ or $b = \pm1$. Prime is defined as follows: an integer $n$ is prime iff whenever $a,b \in \mathbb{Z} $ and $p\mid ab$ then it must be true that $p\mid a$ or $p\mid b$ (or both).    Hint: Prove that if $n\mid ab$ but $n$ does not divide $a$, then $\gcd(a,n) = \pm1$. Then use Bézout's theorem and prove $n\mid b$.
11582
Is my proof beneath perfect and complete?  I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: $$\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.$$  I start by assuming that $m^{\frac 1n}$ is rational and non-integer. So there exist  co-prime integers $a,b$ so that $$\sqrt[n]{m}=\frac{a}{b}$$ $$\implies  m=\frac{a^n}{b^n}\in\mathbb{N}.$$ But since $a$ and $b$ have no common factor, $a^n$ and $b^n$ also have no common factor. So: $$\frac{a^n}{b^n}\notin\mathbb{N},$$ a contradiction.
119382
Consider $X = (a+b)^n-a^n-b^n$  where  $a$ and $b$ and $a+b$ are not divisible by $n$, $n&gt;=3$ is positive and prime, $a \ne b$ and both are greater than 1.  By the binomial expansion we know $X$ is divisible by $n$, but how can we study whether $X$ is divisible by $n^2$ or other powers of $n$? Specifically I am interested to prove $X$ is NOT divisible by $n^2$, but I am not sure how to prove or disprove it.
38532
I'm trying to understand the following proof of why      No two Fermat numbers have a common divisor greater than $1$.   Hardy and Wright show the following proofs:   If $x = 2^{2^n}$, we have  $$\frac{F_{n+k} - 2}{F_n}(\circ) = \frac{2^{2^{n+k}}-1}{2^{2^n}+1} = \frac{x^{2^k}-1}{x+1} = x^{2^k-1} - x^{2^k-2} + \ldots -1$$  The main issue I'm having is the $\circ$ part. I don't understand how they get that. The rest of the proof makes sense but the first part is really confusing me. Can someone please explain where they get that fraction?  Thanks a bunch!
37864
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.   $(a,b)=1$ means $a$ and $b$ have no prime factors in common  $ab$ is simply the product of factors of $a$ and factors of $b$.  Let's say $k\mid a+b$ where $k$ is some factor of $a$.  Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.  So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].  But $(a,b)=1$ so $a$ cannot divide $b$.  We have a similar argument for $b$.  So $a+b$ is not divisible by any factors of $ab$.  Therefore, $(a+b, ab)=1$.  Would this be correct? Am I missing anything?
96129
I know how to prove that, if $2^n-1$ is prime and $n&gt;1$, then $n$ is prime.  But how do we prove that, if $a^n-1$ is prime and $n&gt;1$, then $a$ must equal 2?
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I am reading a math book. It states the following rule:      For an integer $n$ greater than 1, let the prime factorization of $n$ be $$n=p^a_1p^b_2p^c_3...p^m_k$$    Here $a, b, c,..., m$ are nonnegative integers, and $p_1,p_1,...,p_k$ are prime numbers. Then the number of divisors of $n$ is: $$d(n)=(a+1)(b+1)(c+1)...(m+1)$$   I can use the rule to solve problems correctly, but I don't understand the rule. How to prove it?
105980
Let $a$ and $b$ be integers. Show that if $a \equiv b \pmod{p}$ for every prime $p$, then $a = b$. My attempt: Suppose that $p\mid (a-b)$ and $a-b\neq 0$. Then $p\leq|a-b| $. Since the set of prime numbers is infinite we can take $p'$ such that $p\leq |a-b|&lt;p'$. But $p'$ divides $a-b$ as well, thus $p'\leq|a-b|$, which is a contradiction. So $a-b=0$, that is, $a=b$. I am right?
82492
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 $a\mid n$ and $b\mid n$. Prove that $ab\mid n$   Proof(so far):   $\gcd(a,b)=1$ implies $am+bk = 1$ for some integers $m$ and $k$.   (Given hint) Multiplying the equation by $n$ yields: $$amn+bkn=n$$  I'm not sure what to do after applying the hint. Am I supposed to show that $ab$ divides both $amn$ and $bkn$?  Thanks in advance.
4839
We can prove that there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:     Is it true that for every pair of primes $p&lt;q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p&lt;q$ and an integer $n&gt;p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n&gt;6$ then $2^n-1\nmid 5^n-1$?   Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &amp;3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &amp;7^{4}-1\mid 151^{4}-1. \end{array}  Now I prove there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1.$  Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.  If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again.
99272
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.    My attempt:  $p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$ So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.    I do not know what to do next. Please help. Thank you.
35496
Let N denote the number, and suppose $N=a^p \times b^q \times c^r \cdots$, where $a,b,c,\cdots$, are different prime numbers and $p,q,r,\cdots$ are positive integers.Then it is clear that each term of the product   $$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots$$  is a divisor of the given number,and that no other number is a divisor; then how could we use this to show/find that the number of ways in which this $N$ can be resolved into two factors which are prime to each other is $2^{n-1}$,where $n$ is the number of different prime factors in $N$?
36335
I need to show that if $(n-2)! \equiv 1 \pmod n$ then $n$ is prime. I think that if $n$ is composite, then we'll have every factor of $n$ in $(n-2)!$, and it would yield that $(n-2)! \equiv 0 \pmod n$. However, I didn't use the fact that it specifically congruent to $1 \bmod n$, so I think I'm getting something fundamental wrong. Is my solution correct? Why do we demand congruence to $1 \bmod n$?
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$a\in \mathbb{N}$, $p$ prime, $a&lt;p$ prove that $a\mid p+1\iff \exists\, b,c\in\Bbb N:\dfrac{a}{p}=\dfrac{1}{b}+\dfrac{1}{c}$ my attempt: $a\cdot b \cdot c=p\cdot(b+c)$ . i dont know how to use the given $a\mid p+1$
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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 $a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer? But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?  I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.  Are they just a coincidence or is there a theory behind?  Is it about modular arithmetic?
42349
I would like to show that $a^n \mid b^n$ implies $a \mid b$  I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless $m$ is prime. Is that right?  So I am not sure how to approach this one. Any ideas?  Thanks!