Is there a proof of the irrationality of $\sqrt{2}$ that involves modular arithmetic?

Question

I was reading Ian Stewart's Concepts of Modern Mathematics.

Using congruences, It's possible to explain why all perfect squares end in $0,1,4,5,6,9$ but not in $2,3,7,8$.

With this I had the idea of exploring the congruences for both sides of $n^2=2m^2$ in Mathematica:

Table[Mod[n^2, 9], {n, 0, 20}]

Table[Mod[2 m^2, 9], {n, 0, 20}]

And had the results:

{0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4}

{0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8}

But I'm still not sure if the outputs really show what I'm looking for, I have also tried $mod \;10$. The idea is still pretty loose in my mind, I'm stuck on deciding if this proves something or what directions I could take in this enterprise.

Answer 1

$x^2-2$ is irreducible over $\mathbb{Z}$ by reduction since it is irreducible over $\mathbb{F}_3$. Check directly $0^2-2 \equiv 1, 1^2-2 \equiv 2, 2^2-2 \equiv 2 \bmod 3$.

Another alternative: If $z \in \mathbb{Z}$ with $z^2=2$, then the $2$-adic valuation gives $2 \cdot v_2(z)=v_2(2)=1$, contradiction.

Answer 2

This is just the standard proof, rewritten in modular arithmetic:

The key here is that $\gcd(m,n)=1$.

Now, look at $$n^2=2m^2 \pmod{4},$$ in all three cases:

• $m,n$ both odd.
• $m$ even, $n$ odd
• $m$ odd, $n$ even

This proof is pretty artificial though.