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(a) For all positive integers $n$, we have gcd$(2n−1, n) = 1$.

For this, first I tried the Euclidean Algorithm. I divided $2n-1$ by $n$, and got $n-(1/n)$. Then, I didn't know what to do with $1/n$.

I also tried expressing them by a multiple. Say the gcd of them is $d$, so $2n-1=dk$ and $n=dl$ for natural numbers $k,l$. I eventually arrived at $d(2l-k)=1$, and didn't know what to do.

(b) For all positive integers $n$, we have gcd$(4n−2, n) = 2$.

*(b) is solved.

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closed as off-topic by Aqua, Joffan, JMoravitz, Erick Wong, Namaste Nov 7 '17 at 1:12

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Hint: Suppose $d$ is simultaneously a divisor of $2n-1$ as well as $n$. Then this means that $2n-1=kd$ for some integer $k$ as well as..... $\endgroup$ – JMoravitz Nov 6 '17 at 18:32
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    $\begingroup$ Please show what progress you have made towards an answer, so people will know what level of response will be helpful. $\endgroup$ – Joffan Nov 6 '17 at 18:32
  • $\begingroup$ @Joffan sorry, I've added details in my edit $\endgroup$ – tigerustin Nov 6 '17 at 18:41
  • $\begingroup$ You arrived at $d(2l-k)=1$ and so $d$ is a divisor of $1$. What are the possible divisors of $1$? What does that mean $d$ is equal to? What does that mean $\gcd(2n-1,n)$ is equal to? $\endgroup$ – JMoravitz Nov 6 '17 at 19:09
  • $\begingroup$ @JMoravitz We know $d(2l-k)=1$, so $2l-k=1/d$. We know $l, k, d$ are natural numbers so $d$ must be 1. Correct? $\endgroup$ – tigerustin Nov 6 '17 at 19:18
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b) is wrong, take $n=3$ then we get $$\gcd(10;3)=1$$

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    $\begingroup$ Why not just $n=1$? (i.e. OP did absolutely no work to test the hypothesis) $\endgroup$ – Erick Wong Nov 6 '17 at 18:36
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    $\begingroup$ this better, and simplier, but i like the number $3$ $\endgroup$ – Dr. Sonnhard Graubner Nov 6 '17 at 18:37
  • $\begingroup$ It is a great number :). $\endgroup$ – Erick Wong Nov 6 '17 at 18:37
  • $\begingroup$ a prime number and a Erdos number $\endgroup$ – Dr. Sonnhard Graubner Nov 6 '17 at 18:38
  • $\begingroup$ My favorite use of $3$ has to be its position in front of all other naturals in the total ordering established by Sharkovskii’s theorem. $\endgroup$ – Erick Wong Nov 6 '17 at 18:41
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For non-zero integers $a,b$ and integer $c$ we have $$\gcd (a,b)=\gcd (a-bc,b).$$ Proof: For integer $d\ne 0$ we have

(i). If $d|a$ and $d|b$ then $d|(a-bc)$. Because $a/d,b/d\in \Bbb Z$ so $(a-bc)/d=(a/d)-(b/d)c\in \Bbb Z.$

(ii).If $d|(a-bc)$ and $d|b$ then $d|a$. Because $(a-bc)/d, b/d\in \Bbb Z$ so $a/d=((a-bc)+bc)/d=(a-bc)/d+(b/d)c\in \Bbb Z.$

With $a=2n-1, b=n, c=2$ we have $\gcd (2n-1,n)=\gcd ((2n-1)-2n, n)=\gcd (-1,n).$ The only divisors of $-1$ are $\pm 1$ and the larger divisor $+1$ divides every $n\in \Bbb Z.$ So $\gcd(-1,n)=1.$

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For a), $\gcd(2n-1,n) = \gcd((2n-1-n),n) = \gcd(n-1,n) = 1$ for any consecutive numbers.

For b), $\gcd(4n-2,n) = \gcd(4n-2-3n,n) = \gcd(n-2,n) = \gcd(n,2)$. This could be either $2$ or $1$, depending on the parity of $n$.

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