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Let $a,b \in \mathbb Z$. I need to calculate the $\mathrm{gcd}$ of a linear combinations of a,b knowing $\mathrm{gcd}(a,b)$.

I know that $a^2 + b^2 = 0 (\mathrm{mod} 4)$ and $5a+7b=2$, I used this info to find that $\mathrm{gcd}(a,b) = 2$ now I am asked to calculate $\mathrm{gcd}(7a+14b, 14a+21b)$. This is as far as I got:

$\mathrm{gcd}(7a+14b, 14a+21b)$ = $\mathrm{gcd}(7(a+2b), 7(2a+3b))$ = $7\cdot\mathrm{gcd}(a+2b, 2a+3b)$

Lior B
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    You can subtract the first entry of the $\operatorname{gcd}$ twice from the second to get $=7\operatorname{gcd}(a+2b,-b)$. Then add the second twice to the first to get $=7\operatorname{gcd}(a,-b)$. Multiplying by $-1$, an invertible element, doesn't matter, $=7\operatorname{gcd}(a,b)$. – conditionalMethod Nov 14 '19 at 19:23
  • The Euclidean algorithm is your friend. For now. (It doesn't always work on polynomials with integer coefficients, not even linear ones). – Arthur Nov 14 '19 at 19:23

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