Prove that for every natural numbers $a$ and $b$, there are infinitely many numbers $s$, such $\gcd(a+s,b+s)=1$ and $a\neq b$
I tried to use Bezout's theorem but I can't get to the result
As rightfully pointed out in the other answer, the statement trivially does not hold when $a = b$. Thus we must make the assumption that $a \neq b$. Then, my original answer holds:
Without loss of generality, $a > b$. A property of the greatest common divisor tells us that $$ \gcd(a + s, b + s) = \gcd((a + s) - (b + s), b + s) = \gcd(a - b, b + s). $$ Hence, if $p$ is any prime that does not divide $a - b$, we can choose $s = p - b$, and then $$ \gcd(a + s, b + s) = \gcd(a - b, p) = 1. $$ Since there are infinitely many primes, this shows that there are infinitely many such $s$.
As stated the claim is obviously false: $a$ and $b$ may be equal, in which case the GCD is $|a+s|$ (assuming you meant integer $s$, otherwise the problem doesn't make sense), and that is equal to $1$ for exactly two values of $s$.
So you must also have the assumption that $a \ne b$. Let's assume $a < b$.
There are infinitely many primes greater than $b$. For any such prime $p$, let $s = p - b$. Then $\gcd(a+s,b+s)=\gcd(a+p-b, p) = 1$, because $p$ is prime and $ 1 < a+p-b < p$