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？

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.
