Note, In this question: $(a,b)$ means gcd $(a,b)$
Given $(a,b)=2$ Plus, $5a+7b=2$
I need to calculate: $(7a+14b, 14a+21b)$
I did the following:
$$(7a+14b, 14a+21b) = (7(a+2b), 7(2a+3b)) = 7(a+2b, 2a+3b) = (4-3a, 6-a)$$
but stuck here...
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4$7(a+2b, 2a+3b)=7(a+2b, a+b)=7(b,a+b)=7(b,a)=…$ – Macavity Nov 01 '21 at 03:59
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@Macavity you didn't explain anything... which didn't help at all – Roy Nov 01 '21 at 06:00
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2All you need is $\gcd(x,y)=\gcd(x, y\pm x)$. Once you show that, the above chain of equations will make sense. – Macavity Nov 01 '21 at 06:59
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$(a+2b,2a+3b) = (a,b)$ by the linked dupe. – Bill Dubuque Nov 01 '21 at 07:48
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See also the many "Linked" posts in the dupe for other methods. – Bill Dubuque Nov 01 '21 at 07:56
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@Macavity and how to prove this? – Roy Nov 01 '21 at 11:57
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@Roy Try first principles - use the definition of gcd, should be enough to get this done. – Macavity Nov 01 '21 at 13:14
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Assume $a=2p$ and $b=2q$ where $p$ and $q$ are coprime integers. Hence we get $5p+7q=1$. It can be shown that $p=3$ and $q=-2$ is a solution to this diophantine equation, thus a parametric form of the general solution is
$$\begin{cases}p=3+7t\\q=-2-5t\end{cases}$$
And that $\gcd(a+2b, 2a+3b)=2\gcd(p+2q, 2p+3q)=2\gcd(1+3t, t)$.
Since $\gcd(3t, t) = t$, it can be shown that $\gcd(1+3t, t) = 1$.
Hence $\gcd(a+2b, 2a+3b)=2$, and $7\gcd(a+2b, 2a+3b)=14$.
