On prime factors with $n^2+n+1$ 
Show that: There are infinitely many positive integers $n$  such that all prime divisors of $n^2+n+1$ are not  greater than $c\cdot n^{0.8}$, where $c$ is constant.

Maybe this $0.8$ is not best constant?  can you find the smaller number?  This is (Peking University mathematics competition)
 A: In fact one can take $0.8$ arbitrarily small!  Let $\alpha>0$.  Then since $\prod_{p\text{ prime}} \big(1-\tfrac1p\big) \to 0$ one can choose an integer $r$ composed of many small primes such that $\tfrac{\phi(r)}{r} \le \tfrac\alpha2$, and thus $\phi(3r) \le 2\phi(r) \le \alpha r$.
If we choose $n = m^r$ for some $m > 1$, then $n^3 - 1 = m^{3r}-1$ factors into the product of cyclotomic polynomials $\Phi_d(m)$ for each $d$ dividing $3r$.  Each polynomial $\Phi_d$ has degree $\phi(d)$ which divides $\phi(3r) \le \alpha r$, so each factor $\Phi_d(m)$ has size at most $C m^{\alpha r}$, where $C$ depends on the coefficients of all the cyclotomic polynomials $\Phi_d$.  Thus any prime factor of $n^3-1 = (n^2+n+1)(n-1)$ can be no larger than $Cm^{\alpha r} = Cn^\alpha$.
For $\alpha = 0.8$ we can simply choose $r = 10$ so that $\phi(3r) = 8 = \alpha r$.  Looking at the cyclotomic polynomials up to $30$ shows we may comfortably take $C=5$ (in general we may use $C = 1+o(1)$ by taking $m$ large enough, since $\Phi_d$ is monic).
