The locker problem - why squares?

Squares are the only integers which have an odd number of divisors.

All other (non-square) integers have an even number of divisors.

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A deeper insight:

Given an integer $n$, for every integer $d$ which divides $n$, the integer $n/d$ also divides $n$.

If $n$ is non-square, then for every integer $d$ which divides $n$, the integer $n/d \neq d$.

So we can split the divisors of $n$ into pairs, hence $n$ has an even number of divisors.

If $n$ is square, then for every integer $d\neq\sqrt{n}$ which divides $n$, the integer $n/d \neq d$.

So we can split the divisors of $n$ except $\sqrt{n}$ into pairs, hence $n$ has an odd number of divisors.

If a locker number has an even number of factors, it will be alternatively opened and closed and even number of times, ending in the same configuration it started.

Square numbers have a odd number of factors.

The idea here is that it is easy to identify pairs of students who will open and close a locker (well, as easy as any interesting math problem ever is!). If any given student will open or close a locker, it means their number is a divisor of the locker's number.

Consider locker X. Let's assume there is a student Y who toggles the locker. This means we know X/Y is an integer. This also means that student X/Y will also toggle the locker.

Take locker 12, with factors 1, 2, 3, 4, 6, and 12. If I rearrange them into pairs, (1, 12) (2, 6) (3, 4) we can see that for every student who opens a locker, there will be one that closes the locker.

Now, Let's look at a square. Let's take 16, with factors 1, 2, 4, 8, 16. If I rearrange them into pairs I get (1, 16) (2, 8) (4, 4)... but wait... I just had to use 4 twice. I had to do that because it was a square. However, student 4 is only going to toggle locker 16 once. Thus, the locker will be left open at the end of the day!