Team Selection Test

By induction over $k$ we will prove that for each pair of positive integers $(a,b)$ there is a positive integer $n$ such that $n^2+an+b$ has at least $k$ distinct prime divisors. The case $k=1$ is obvious. For given $(a,b)$ suppose that for some $n_k$, $n_k^2+a{n}_k+b$ has $k$ distinct prime divisors $p_1,p_2,\dots ,p_k$. Then $p_1{p}_2\dots p_k$ divides $n_k^2+a{n}_k+b=m$. Let us define $n_{k+1}=n_k(m^2+1)$. Then $$n_{k+1}^2+a{n}_{k+1}+b=n_k^2(m^2+1{)}^2+a{n}_k(m^2+1)+b$$ $$=n_k^2(m^4+2m^2+1)+a{n}_k(m^2+1)+b=n_k^2{m}^4+2m^2{n}_k^2+a{n}_k{m}^2+m$$ $$=m(n_k^2{m}^3+2m{n}_k^2+a{n}_{k}m+1)$$ The last expression is divisible by $p_1,p_2,\dots ,p_k$ and some distinct prime number $p_{k+1}$ since $m$ and $n_k^2{m}^3+2m{n}_k^2+a{n}_{k}m+1$ are relatively prime. Done.