That proof is really great! It&#x27;s always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles.<p>However, I slightly disagree with the introduction of the proof:<p><i>| The really interesting thing about this proof is that it doesn&#x27;t use divisibility, just mathematical induction in its &quot;Z is well-ordered&quot; form.</i><p>That&#x27;s quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility.<p>However, I don&#x27;t see that. (Does anyone else see how this proof could work on such a generalized setting?)<p>Instead, it seems that this proof introduces the concept of &quot;divisibility&quot; through the backdoor, by calculating with fractions and making use of the basic fraction laws, which are based on divisibility. In particular, the equation<p><pre><code>    1 &#x2F; (sqrt(m) - n) = (sqrt(m) + n) &#x2F; (m - n^2)
</code></pre>
makes use of the fact that you can extend and cancel fractions. Moreover, these fractions are not even fractions in Z, but in R, or at least in Z[sqrt(m)].
So there&#x27;s still a lot of specialized structure involved in this proof.<p>Nevertheless, great proof!