# Prove that there exist $135$ consecutive positive integers so that the $n$th least is divisible by a perfect $n$th power greater than $1$

Prove that there exist 135 consecutive positive integers so that the second least is divisible by a perfect square $$> 1$$, the third least is divisible by a perfect cube $$> 1$$, the fourth least is divisible by a perfect fourth power $$> 1$$, and so on.

How should I go about doing this?

I thought perhaps I should use Fermat's little theorem, or its corollary?

Thanks!

• Wow, this problem sounds painful. Mar 19, 2019 at 4:08

Use the Chinese Remainder Theorem. Pick $$134$$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $$N$$ is the least of your $$135$$ numbers, you have $$N+1 \equiv 0 \pmod {2^2}, N+2 \equiv 0 \pmod {3^3}, N+3 \equiv 0 \pmod {5^4}\ldots$$
By the Chinese remainder theorem, there is an integer $$n$$ such that $$n\equiv -k\ (\text{mod}\ p_k^{k+1})$$ for all $$k=1,2,\dots 134$$, (where $$p_k$$ is the $$k^\text{th}$$ smallest prime). Then $$n, n+1, \dots, n+134$$ satisfy the required condition.