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Got across this question and haven't figured out how to solve it:

Given two integers $a,b$ such that

  • $a$ is even.
  • $8a+15b=7$.
  • $a-b=0 \pmod{7}$.

Find the GCD $(2a+4b,2a+8b)$. The answer is 28.

In my attempts I have tried to simplify on variable with the last fact

$a=7n+b$

$8(7n+b) +15b=7$

$2b=0 \pmod 7$

$b=0 \pmod 7$

and after that I have tried to imply Euclid's algorithm to find the GCD with no success.

RobPratt
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cenzor
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    Please edit to include your efforts. Finding some examples of suitable $a,b$ is a good start...if nothing else, that at least tells you the answer (which you would then still have to prove). – lulu May 31 '22 at 17:16
  • It is better style to use \equiv instead of = in a congruence. – DanielWainfleet May 31 '22 at 20:27

1 Answers1

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$a-b\equiv 0\mod 7$ and $b\equiv 0 \mod 7,$ so $a\equiv 0\mod 7.$ And $a$ is even. So let $a=14a'$ and $b=7b'.$

From $8a+15b=7$ we get $16a'+15b'=1$. Note this implies that $a',b'$ are co-prime.

So $GCD(2a+4b,2a+8b)=GCD(28a'+28b',28a'+56b')=28\cdot GCD(a'+b', a'+2b').$

Now GCD$(a'+b', a'+2b')=1.$

Proof: If $p|a'+b'$ and $p|a'+2b'$ then $p|(a'+2b')-(a'+b')=b',$ hence also $p|2b',$ so $p|(a'+2b')-(2b')=a',$ so $p|a'$ and $p|b'.$ But $a',b'$ are co-prime, so $|p|=1.$

DanielWainfleet
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    For your last part, note that, using $\gcd(a, b) = \gcd(a + mb, b) = \gcd(a, b + ma)$ for any integer $m$, we have $\gcd(a' + b', a' + 2b') = \gcd(a' + b', b') = \gcd(a', b') = 1$. – John Omielan May 31 '22 at 20:48
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    @JohnOmielan . Of course... I was merely "walking through" it. – DanielWainfleet Jun 01 '22 at 00:43