there are infinitely many $n$ such that $p(n)Let $p(n)$ denote the largest prime factor of $n$. Prove that there are infinitely many $n$ such that $$p(n)<p(n+1)<p(n+2)<p(n+3)$$
if the three consecutive numbers,I can prove it,also can see
three,erdos,but for four (or more) consecutive number maybe is old？and How do to？
 A: By standard conjectures (see Dickson's conjecture), there exist infinitely many $k$ such that $3k+1$, $20k+7$, $30k+11$, and $60k+23$ are all prime. Then the required inequalities hold for $n=60k+20$, since $p(n)=3k+1$, $p(n+1)=20k+7$, $p(n+2)=30k+11$, and $p(n+3)=60k+23$, and clearly $3k+1<20k+7<30k+11<60k+23$.
I don't know whether there is an unconditional proof. I note that Erdos and Pomerance were unable to prove the existence of infinitely many $n$ such that $p(n)>p(n+1)>p(n+2)$, so this could be a very hard problem.

Original, incorrect, answer follows:
By standard conjectures (see Dickson's conjecture), there exist infinitely many $k$ such that $3k+2$, $10k+7$, $15k+11$, and $30k+23$ are all prime. Then the required inequalities hold for $n=30k+20$, since $p(n)=3k+2$, $p(n+1)=10k+7$, $p(n+2)=15k+11$, and $p(n+3)=30k+23$, and clearly $3k+2<10k+7<15k+11<30k+23$.
I don't know whether there is an unconditional proof. I note that Erdos and Pomerance were unable to prove the existence of infinitely many $n$ such that $p(n)>p(n+1)>p(n+2)$, so this could be a very hard problem.
