Lucy and I found this proof last night. I'm sure it is not new, but hadn't seen it before.
We aim to prove that the square root of is irrational, i.e. that there is no rational such that .
Proof: By contradiction. Suppose that there was such a . Then, as a rational number, it could be written in lowest terms: i.e. as for such that and are relatively prime.
Consider the parities of and . There are four options. Note that the parities of and determine the parities of and .
even, even. This cannot happen since then and would share the factor , violating our assumption that they were relatively prime.
even, odd. This cannot happen since an even number divided by an odd number is always even, but is odd.
odd, even. This cannot happen since an odd number divided by an even number is not an integer, but is an integer.
odd, odd. This possibility is the only one that is not ruled out immediately. But it too cannot happen. Write and , where . Then:
Now consider the equation mod . The left side is equal to , but the right side is equal to , so we have a contradiction.
Notes: Consider using this proof to prove that the square root of is irrational. We also conclude that both and must be odd, but since the last bit does not work. And indeed it better not, otherwise we could prove that the square root of was irrational.
This inelegant, bedraggled proof applies unevenly and will not win any beauty contests.