It is true that the pairs \((5x+3y,3x+2y)\) and \((x,y)\) have the same GCD. Let’s first see how to find a relationship between \(a\) and \(b\) if we want to find the \(gcd(a,b)\).
\[gcd(a,b) = gcd(amodb,b)\]
Let \(x = amodb\), and let k be any number.
\[x-a = bk\] \[x=a+bk\]
Let \(k=-1\)
\[x=a-b\]
Then, subsitute x back:
\[gcd(a,b) = gcd(a-b,b)\]
\[gcd(5x+3y,3x+2y) = gcd(2x+y,3x+2y)\]
\[gcd(2x+y,3x+2y) = gcd(x+y,2x+y)\]
\[gcd(x+y,2x+y) = gcd(x,x+y)\]
\[gcd(x,x+y) = gcd(x,y)\]
Using the above reasoning for finding GCD, we ahve to pick a clever \(k\) that is an integer and will make \(a\) reduce to 1 in \((a,b)\):
\[(S_{n} - k S_{x}, S_{x})\]
Let:
\[k= \frac{S_{n} - 1}{S_{x}}\]
To prove k is an integer, for x < n:
\[\frac{S_{n} - 1}{S_{x}}= \frac{S_{i}*S_{i-1}*S_{i-2}...S_{x}*S_{0}}{S_{x}}\]
\(S_{x}\) in the numerator and denominator will cancel out, and prove it is an integer.
So:
\[gcd(S_{n} - \frac{S_{n-1}}{S_{x}}*S_{x},S_{x}) = gcd(S_{n} - (S_{n} -1), S_{x})\] \[=gcd(1,S_{x})=1\]
Since the gcd is 1, any two numbers in the sequence, is relatively prime.