Talk:Fermat's Two Squares Theorem
Proof of uniqueness :
With the same technique, there exists the following faster and brighter way:
Assuming $p=a^2+b^2=c^2+d^2$
1) $(ac+bd)(ad+bc)=(a^2+b^2)cd+(c^2+d^2)ab=p(ab+cd)$
This implies (for example) $p \mid (ac+bd)$ and thus $ac+bd\geq p$ (permuting $c$ and $d$ would end up similarly).
2) $p^2=(ac+bd)^2+(ad-bc)^2\geq p^2+(ad-bc)^2$ thus $ad-bc=0$
So $\dfrac{c^2}{a^2}=\dfrac{d^2}{b^2}=\dfrac{c^2+d^2}{a^2+b^2}=1$
Hence $c=a$, $b=d$
Maybe this result should also be published in the page?
- I'm struggling to see why $a d - b c = 0$ implies $\dfrac {c^2 + d^2} {a^2 + b^2} = 1$ necessarily.
- Yes, we have that $a d - b c = 0$ implies that $\dfrac {c^2} {a^2} = \dfrac {d^2} {b^2}$, but it does not then necessarily follow that $\dfrac {c^2 + d^2} {a^2 + b^2} = 1$.
- If $c = 2$ and $a = 1$ and $d = 4$ and $b = 2$ then $\dfrac {c^2} {a^2} = \dfrac {d^2} {b^2}$ indeed, but $\dfrac {c^2 + d^2} {a^2 + b^2} = \dfrac {20} {5} = 4$. --prime mover (talk) 13:00, 14 September 2023 (UTC)
14:48, 14 September 2023 (UTC)
Hello prime mover
That's the initial hypothesis on $p$ that makes the ratio equal 1
$\dfrac{c^2+d^2}{a^2+b^2}=\dfrac{p}{p}=1$
- oh yeah suppose so --prime mover (talk) 15:30, 14 September 2023 (UTC)
18:08, 15 September 2023 (UTC):
Hello [[User:Prime.mover|: why did you remove the proof of uniqueness chapter in the page?
- i didnt --prime mover (talk) 18:37, 15 September 2023 (UTC)
19:10, 15 September 2023 (UTC)
OK. I've just understood that you've turned "Uniqueness Lemma" title into a link. Not very clear that it contains the proofs, indded but there are no removed.
Now I have remarks on your newly created page, because Usagiop has added a 3rd version that is actually just a modified version of the first.
I will discuss that on the new page.
- yeah dunno y he did that, ask him --prime mover (talk) 19:37, 15 September 2023 (UTC)