Is there a limit to how many integer-valued strictly increasing quadratic functions $f(x):=ax^2+bx+c$ can be guaranteed to yield coprime values for a given $x\in\mathbb N$?
For example, the values of
$$x^2+x+1 \\ x^2+x+5 \\ 2x^2+2x+7$$
are always mutually coprime. I've found I can't do much better than about a dozen of these before it seems impossible to add another one, so I'm wondering if there's a good reason behind it.
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size of factors affects things, your first polynomial will not be coprime with all integer inputs $x$ into $1\over 3$ of all integer coefficient, integer value polynomials. This happens because: $$3\mid (3n+r)x^2+(3m+r)x+(3p+r),\quad x\equiv 1\bmod 3, n,m,r,p\in\mathbb{N}$$ along with 6 others out of the 27 mod 3 polynomials, I couldn't be bothered to type out. like in integer arithmetic 3 divides a third of the polynomials at $x=1$, 5 a fifth, etc. in your case we get 3,7,11 dividing $37\over 77$ of all polynomials. just shy of half. Your only hope then, is they don't overlap in the $x$ values that work.
With quadratic and even with linear polynomials we can get an arbitrarily large set.
For example, take the sequence $$ 50! x^2 + 53, 50! x^2 + 59, 50! x^2 + 61, \dots, 50! x^2 + 89, 50! x^2 + 97. $$ Here, I began by choosing all the prime numbers in the range $[50,100]$ as the constant terms. Their differences are all numbers in the range $[0,50]$.
So I added $50! x^2$ to each polynomial (I could have added $50! x$). Since $50!x^2$ is always divisible by primes in the range $[2,50]$ but the constant term never is, none of these polynomials are ever divisible by any of these small primes, and so their GCDs are always $1$.
In general, for any $k$, we can take the polynomials $k! x^2 + p$ for prime numbers $p$ in the range $(k,2k]$, and they will be relatively prime at any natural number $x$ for a similar reason. The number of such primes grows without bound, and therefore so does the number of polynomials we get.