# 4 Consecutive Integers cannot be Square-Free

Jump to navigation
Jump to search

## Theorem

Let $n, n + 1, n + 2, n + 3$ be four consecutive positive integers.

At least one of these is not square-free.

## Proof

Exactly one of $n, n + 1, n + 2, n + 3$ is divisible by $4 = 2^2$.

Thus, by definition, one of these is not square-free.

$\blacksquare$

## Sources

- 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $29$