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For the following keep in mind that by (a,b) i mean the gcd of a,b

Let the gcd of $(a,b)=1$

IF $d|(a+b)$ prove that $(a,d)=(b,d)=1$

As an analysis student this is my first time approaching algebra and i thought it would be the best to start by some basic number theory exercises but im struggling with this one could someone help me take on this one?

Arturo Magidin
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nick hanzo
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    Well, suppose otherwise. Let, say, $p$ be a prime dividing $\gcd(a,d)$. Derive a contradiction. – lulu Jul 18 '21 at 17:37
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    Re the comment of @lulu, you need the intermediate result that in general, if $r$ divides $s$ and $r$ divides $t$, then $r$ divides $(s + t)$. Similarly, if $r$ divides $s$ and $r$ divides $(s + t)$, then $r$ divides $t$. You should be able to prove these intermediate results by using the idea that $r$ divides $s$ if and only if there exists an integer $k$ such that $(r \times k) = s.$ – user2661923 Jul 18 '21 at 17:42
  • @user2661923 thanks for your replies i believe i need to to do some more studying first before i take on this one. Your answers should be proven helpfull later on :P – nick hanzo Jul 18 '21 at 17:51
  • $b = -a+qd \Rightarrow (d,b) = (d,-a+qd) = (d,-a) = (d,a),$ by the linked dupe. – Bill Dubuque Jul 19 '21 at 00:58

1 Answers1

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Assume the opposite. Let gcd(a,d) > 1.

Then, there exists some prime p that divides a and d. Let a = mp, d = np, where m, n are integers.

Next, consider d divides a + b. This means there exists some integer s such that a+b = sd.

Substituting, we have mp + b = snp. So b= snp - mp. The right hand side is clearly divisible by p, so b is divisible by p. Thus both a and b are divisible by p, which is a contradiction, since gcd(a,b) is given to be 1.

Since there is a contradiction, it is impossible gcd(a,d) > 1, so therefore gcd(a,d) = 1.

The proof for gcd(b,d) > 1 is the same (just swap all instances of a and b in the previous proof).

abnew123
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  • Doesn't $\gcd(a,b) = p$ say that $p$ divides $a$ and also $p$ divides $b$? $\gcd$ is a abbreviation of greatest common divisor, meaning that $p$ divides $a$ and also $p$ divides $b$ and no number larger than $p$ could possibly divide both $a$ and $b$. So, how does "Thus, both $a$ and $b$ are divisible by $p$ which is a contradiction since $gcd(a,b)=p$" make sense? – Dilip Sarwate Jul 18 '21 at 18:53
  • Yep you're right, I meant to say gcd(a,b) is 1, not p. – abnew123 Jul 18 '21 at 19:02
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    Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 19 '21 at 00:59