<s>[INST] greatest common divisor of two elements Find all possible values of GCD(4n + 4, 6n + 3) for naturals n 
and prove that there are no others
3·(4n + 4) - 2·(6n + 3) = 6, 
whence the desired GCD is a divisor 6. 
But 6n + 3 is odd, so only 1 and 3 remain.
n=1 and n=2 are examples for GCD=1 and GCD=3
is the solution correct ?
any other way to solve this ?
 [/INST] Your way is correct.
Other way. 
$\gcd(4n+4, 6n+3) = \gcd(4n+4, (6n+3) - (4n+4)) =$
$\gcd (4n+4, 2n -1) = \gcd(4n+4 - 2(2n-1), 2n-1)=$
$\gcd (6, 2n- 1) = $
... Now two things should be apparent.  $2n-1$ is odd and $6$ is even so the prime factor  $2$ of $6$ will not be a factor of $2n-1$.  And Lemma: if $\gcd(j,b) = 1$ then $\gcd(j*a, b) = \gcd(a,b)$.  That can be easily proven many ways.
So $\gcd(2*3, 2n-1) = \gcd(3,2n-1)$.  Which is equal to $3$ if $3|2n-1$ which can happen if $2n-1 \equiv 0 \pmod 3$ or $n\equiv 2 \pmod 3$.  Or is equal to $1$ if $3\not \mid 2n-1$ which can happen if $n\equiv 0, 1 \pmod 3$.
And another way:
$\gcd(4n+4, 6n+3) = \gcd(4(n+1), 3(2n+1)=$.
... as $3(2n+1)$ is odd....
$\gcd(n+1, 3(2n+1))$.
Now $\gcd(n+1, 2n+1) = \gcd(n+1, (2n+1)-(n+1) = \gcd(n+1, n) = \gcd(n+1 - n, n) = \gcd(1, n) = 1$.
So... $\gcd(n+1, 3(2n+1)) = \gcd(n+1, 3)$.
Which is $3$ if $3|n+1$ and is $1$ if not.
Perhaps we can retrofit this as
$\gcd(3,n+1) = \{1,3\}$
$\gcd(2n+1, n+1) = 1$ so 
$\gcd(3(2n+1), n+1) = \gcd(3,n+1)$.
$\gcd(3(2n+1), 2) = 1$ so
$\gcd(3(2n+1), 2^2(n+1)) = \gcd(3,n+1)$.
All comes down to "casting out" relatively prime factors.
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