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How do I rather quickly prove that $$a=2500137802, b=1420515313, c=3920653117$$ are relatively prime numbers? I know it has something to do with Euclidean algorithm, but still doesn't ring a bell to me.

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    Yes, this method I know, but it is not a rather quick proof and I would prefer someone showing me how to do this with reference to Euclidean algorithm. Thanks anyways. – 2017-02-16
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    The Euclidean principle is that $\gcd(a,\, b) =\gcd(a,\, b-a)=\gcd(a,\, b-\lfloor{b/a} \rfloor a)$: so you reduce $b$, then invert the couple and reduce $a$, etc. – 2017-02-16

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Hint $\ $ Any common divisor divides $\,2\, =\, 3920653117 -2500137802 - 1420515313$

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    Yeah, I got that, but I don't really know how to make use of it. – 2017-02-16
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    @John Are you trying to show that all three have no common divisor, or that they are pairwise coprime? – 2017-02-16
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    I'm trying to show that all three don't have a common divisor. – 2017-02-16
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    @John Any common divisor divides the RHS above so it divides the LHS $= 2\,$ so ..... – 2017-02-16
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    It's becoming more clear for me now. I'm just learning this stuff, therefore it's really hard to understand everything, even though it looks simple and i feel daft, but thanks. :D – 2017-02-16
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    @John Suppose $\,d\,$ is a common divisor of $\,a,b,c\,$ Then $d$ also divides all integral linear combinations of them, i.e. $\,d\mid ja+kb+nc\,$ for all integers $\,j,k,n.\,$ The Euclidean algorithm is one efficient way to search for the *least* positive such linear combination, which is easily seen to be their gcd. The above exercise was constructed to help us quickly notice a very small linear combination $= 2.\ \ $ – 2017-02-16