Prove the Sequence of Inequalities 
Let $\text{gpd} (n)$ denotes the greatest prime divisor of $n$. Then examine there exist infinitely many $n$ for which we have $\text{gpd} (n)$ $>$ $\text{gpd} (n+1)$ $>$ $\text{gpd} (n+2)$.
More generally find the values of $k$ such that $\text{gpd} (n)$ $>$ $\text{gpd} (n+1)$ $>$ $\text{gpd} (n+2)$ $>$ $\cdots$ $>$ $\text{gpd} (n+k)$.

I cannot find the trick to tackle this problem. It can be shown that for infinitely many $n$ we have $\text{gpd} (n)$ $>$ $\text{gpd} (n+1)$. This is trivial but after this I am unable to make any progress.
 A: Let $ p>2$ be a prime, and let $\displaystyle n_k=p^{2^k}-1$. We have         $\displaystyle P\left(n_k+1\right)=p$. We look at the sequence $\displaystyle  \{n_k+2\}_{\{k\ge 1\}}$. We can see that any 2 distinct terms have 2 as their greatest common divisor, because for $ i<j$ we have $\displaystyle \left(p^{2^j}+1\right)-2 = \left(p-1\right)\left(p+1\right)\left(p^2+1\right) ...\left(p^{2^i}+1\right) ...\left(p^{2^{j-1}}+1\right)$.
Since none of the terms of the sequence considered is divisible by 4, it means that the sequence $\displaystyle P\left(n_k+2\right)$ contains arbitrarily large primes. Let $ k$ be the smallest integer for which $\displaystyle  P\left(n_k+2\right) > p$. Since $\displaystyle n_k=\left(p-1\right)\cdot\left(p+1\right)\cdot n_1\cdot n_2\cdot ...\cdot n_{k-1}$ and all prime divisors of the number on the right are $ <p$, it means that $ P\left(n_k\right)<p=P\left(n_k+1\right)<P\left(n_k+2\right)$, and such a number $ n_k$ can be found for any prime $ p$. Thus there are infinitely many such $n$. I don't think this is original, but I had the idea from the problem to find $\text{gcd}(a^{2^i}+1,a^{2^j}+1)$.
