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Consider a polynomial $f(x)=x^4+2x^3 -2x^2 -4x +4$. Define $$A = \{ f(m) \mid m\in \mathbb{Z}\}$$ And define $$ B =\bigg\{ p \mid p \ {\rm is\ a\ prime\ and}\ \mathbb{Z}\cdot p \bigcap A =\emptyset\bigg\} $$

Hence prove that $|B|=\infty$ i.e. there are infinitely many primes s.t. any multiple is not in $A$.

Proof : If not, there is $C>0$ s.t. a prime $p>C$ satisfies $f(m)=Kp$ for some $m$.

We can assume that $kp\leq m <(k+1)p$. Hence there is $0<i<p$ s.t. $f(i)$ is divisible by $p$.

$$f(x)=x^2(x+3)(x-1)+(x-2)^2$$ so that $$f(0)=4,\ f(1)=1,\ f(2)=20,\ f(p-3)=25$$ are not divisible by $p$.

So how can we prove the non-divisibility for remaining $i$ ?

HK Lee
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    Maybe this uses more prerequisites than are allowed for the contest but you can prove the original claim by noting that up to $N$ there are approximately $N/log(N)$ primes but only about $N^{1/4}$ numbers generated by the polynomial. So asymptotically $A$ is much smaller than $B$ hence there are infinitely many numbers in $B$ but not in $A$. – quarague Aug 23 '21 at 06:00
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    @quarague the question is not about getting primes, but about multiples of primes (the title is a bit confusing) – user8268 Aug 23 '21 at 08:04

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$x^4+2x^3 -2x^2 -4x +4 = (x^2+x-2)^2 + x^2$.

If we wish for this to be $0\bmod p$ then we require that both $x^2$ and its negative be quadratic residues $\bmod p$, since both can't be $0$ we conclude $-1$ must be a quadratic residue, which excludes infinite values of $p$.

Asinomás
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    or, with quadratic form $au^2 + b uv + c v^2$ and discriminant $D$ not a square, if we have an odd prime $q$ that does not divide $D$ and $(D|q) = -1,$ then $q$ cannot divide $au^2 + b uv + c v^2$ unless it divides both $u,v$ What I missed was writing the polynomial as a quadratic form in polynomials. See https://math.stackexchange.com/questions/2291070/prime-divisors-of-k2k12/2291224#2291224 – Will Jagy Aug 23 '21 at 18:30