I found this problem in Discrete Mathematics and Its Applications while studying for an exam and am having trouble solving it. Prove that for $a,b\in N^*, a \ge b, 2\gcd(a,b) = \gcd(a+b, a-b)$ if $a$ and $b$ are odd. My current idea was to use the principles in the Euclidean Algorithm to somehow prove it, but I'm running into a wall. Any help would be great, thanks!
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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:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan
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Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).  Now I have that as $\gcd(b,6) = 1$, we know that  $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)  So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.  Then I'm not sure where to take it from here? Maybe I've gone down the wrong path to start with, and should be looking more along the lines of the Chinese Remainder Theorem? I'm not sure.
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Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.   This is what I have done so far:  Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.  I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.  EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity.
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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 some advice how I can handle this problem? And $a,b,z \in \mathbb{Z}$.
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If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.  My attempt at answering the question:  \begin{align*} x &amp;\equiv 0 \pmod{a}\\\ &amp;\Longrightarrow x\text{ is divisible by $a$}\\\ &amp;\Longrightarrow x = ma\text{ for some integer $m$}\\\ \ \\\ x &amp;\equiv 0 \pmod{b}\\\ &amp;\Longrightarrow  x\text{ is divisible by $b$}\\\ &amp;\Longrightarrow x = mb\text{ for some integer $m$}\\\ \ \\\ x^2 &amp;= (ma)(mb)\\\ x^2 &amp;= (m^2)(ab)\\\ x &amp;= \sqrt{m^2ab}\\\ x &amp;= m\sqrt{a}\sqrt{b} \end{align*} Let $m$ be $k\sqrt{a}\sqrt{b}$. Then \begin{align*} x &amp;= kab\\\ &amp;\Longrightarrow x \equiv 0 \pmod{ab} \end{align*}  Is this correct, if not can someone point me in the right direction?
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I am trying to prove that for every integer $n \ge 1$, there exists uniquely determined $a &gt; 0$ and $b &gt; 0$ such that $n = a^2 b$, where $b$ is squarefree.  I am trying to prove this using the properties of divisibility and GCD only. Is it possible?  Let me assume that $n = a^2 b = a'^2b'$ where $a \ne a'$ and $b \ne b$'. Can we show a contradiction now?
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For the integers $a,b,c$ show that $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$  Proof:  Let $u$ and $v$ be integers. Then $\gcd(a,b)=au+bv$. Then $c\cdot \gcd(a,b)=c\cdot(au+bv)=acu+bcv$ Let $x$ and $y$ be integers. Then $\gcd(a,\gcd(a,b)\cdot c)=ax+(acu+bcv)y=ax+acuy+bcvy=a(x+cuy)+(bc)(vy)$.  Let $d$ and $e$ be integers. Then $\gcd(a,bc)=ad+bce$.   Since $d$ and $e$ are integers let $d=x+cuy$ and $e=vy$. Hence $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$
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Let $p$ be a prime. Suppose that $\gcd(a, b) = p$. Find $\gcd(a^2,b)$  for all integers a and b.   I was able to prove that $\gcd(a^2,b^2) = p$ but I don't know if it helps.  Any help is greatly appreciated.
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I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both $a$ and $b$ to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer.
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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.
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Suppose I know that I can find the $\gcd$ of two integers with the Euclidean algorithm. What is the 'extension' so I can use this method also for polynomials?  Edit: I found an easy answer with less rigor which is approachable without knowledge of matrices:    also the first 3 subchapters of
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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 there...
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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  $a,b,c,z \in \mathbb{Z}$ and $c\neq 0)$
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If integers $a$ and $b$ are not both zero, then $\gcd(a,b)=\gcd(a-b,b)$   Assume either $a=0$ or $b=0$. If $a=0$, then $\gcd(a,b)=\gcd(0,b)=b$ and $\gcd(a-b,b)=\gcd(-b,b)=-b$, so $\gcd(a,b)\neq\gcd(a-b,b)$.  If $b=0$, then $\gcd(a,b)=\gcd(a,0)=0$ and $\gcd(-b,b)=\gcd(a,0)=0$.  So, what can I say? Can you help? Can you check my proof-trying?
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Given that $\gcd(a,b) = 1$ and $\gcd(c,d) = 1$, show that $\gcd(ac,bd) = \gcd(a,d) * \gcd(b,c)$.  The work that I have done so far goes as follows.  We write  $$\gcd(ac,bd) = acv + bdu$$ then  $$\gcd(a,d) = ax + dy\quad\text{and}\quad\gcd(b,c) = bs + ct$$ and so \begin{align} \gcd(a,d) * gcd(b,c) &amp;= (ax + dy) * (bs + ct) \\ &amp;= axbs + axct + dybs + dyct \end{align} I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$  but then I am stuck here. I don't know what else to use to prove the answer.
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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; by these two conclusions how can I conclude $d=1$ or $2$?  And also is there any other way of proving this result?
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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.