# Sequence of 4 Consecutive Square-Free Triplets

## Theorem

The following sets of $4$ consecutive triplets of integers, with one integer between each triplet, are square-free:

$29, 30, 31; 33, 34, 35; 37, 38, 39; 41, 42, 43$
$101, 102, 103; 105, 106, 107; 109, 110, 111; 113, 114, 115$

## Proof

Note that $32, 36, 40$ and $104, 108, 112$ are all divisible by $4 = 2^2$, so are by definition not square-free.

Then inspecting each number in turn:

 $\ds 29$  $\ds$ is prime $\ds 30$ $=$ $\ds 2 \times 3 \times 5$ and so is square-free $\ds 31$  $\ds$ is prime

 $\ds 33$ $=$ $\ds 3 \times 11$ and so is square-free $\ds 34$ $=$ $\ds 2 \times 17$ and so is square-free $\ds 35$ $=$ $\ds 5 \times 7$ and so is square-free

 $\ds 37$  $\ds$ is prime $\ds 38$ $=$ $\ds 2 \times 19$ and so is square-free $\ds 39$ $=$ $\ds 3 \times 13$ and so is square-free

 $\ds 41$  $\ds$ is prime $\ds 42$ $=$ $\ds 2 \times 3 \times 7$ and so is square-free $\ds 43$  $\ds$ is prime

 $\ds 101$  $\ds$ is prime $\ds 102$ $=$ $\ds 2 \times 3 \times 17$ and so is square-free $\ds 103$  $\ds$ is prime

 $\ds 105$ $=$ $\ds 3 \times 5 \times 7$ and so is square-free $\ds 106$ $=$ $\ds 2 \times 53$ and so is square-free $\ds 107$  $\ds$ is prime

 $\ds 109$  $\ds$ is prime $\ds 110$ $=$ $\ds 2 \times 5 \times 11$ and so is square-free $\ds 111$ $=$ $\ds 3 \times 37$ and so is square-free

 $\ds 113$  $\ds$ is prime $\ds 114$ $=$ $\ds 2 \times 3 \times 19$ and so is square-free $\ds 115$ $=$ $\ds 5 \times 27$ and so is square-free

$\blacksquare$