# Biggest prime divisor of $n^4+n^2+1$ and $(n+1)^4+(n+1)^2+1$

Show that there are infinite values of $n$ such that the biggest prime divisor of $n^4+n^2+1$ equals the biggest prime divisor of $(n+1)^4+(n+1)^2+1$.

• They both have a factor $n^2+n+1$. If that is prime and $n^2+3n+3$ is not you have a case. Jun 7, 2017 at 14:45
• @RossMillikan taking $n=3m$ will make sure that $n^2+3n+3$ is composite and $n^2+n+1$ could be prime(i think infinitely many times). Jun 7, 2017 at 15:12
• ${n^2 + n + 1}$ has inf prime ... ? Jun 7, 2017 at 15:30
• @Ahmad: Yes if $n=3m$ you know $9m^2+9m+3$ is composite. I don't know how to justify that $9m^2+3m+1$ is prime infinitely often or I would have answered. Jun 7, 2017 at 15:42

As Ross Millikan commented, both terms contain a factor $$n^2+n+1.$$ What we also note is that in fact $$n^4+n^2+1 = (n^2+n+1)((n-1)^2+(n-1)+1).$$ This motivates definining $$f(n) = n^2+n+1,$$ and we see that \begin{align} f(n^2) = f(n)f(n-1). \tag{1} \end{align} If $$g(n)$$ is the largest prime divisor of $$f(n),$$ we are reduced to showing that $$g(n)$$ is never monotonous for $$n > N$$ for any constant $$N.$$ This is because the condition in the problem statement is equivalent to $$g(n)$$ having a local maximum. But this is clear because of (1), which implies that $$g(n^2) = \min{(g(n),g(n-1))},$$ so if $$g$$ is monotonous for sufficiently large $$n,$$ it must be constant for sufficiently large $$n,$$ which is easily seen to be impossible (for example if it takes on the value $$p,$$ we have $$n^2+n+1 = 0 \pmod p,$$ which can only have at most two solutions $$\pmod p.$$)