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I'm doing the second part of the following exercise in Miles Reid's Undergraduate Commutative Algebra:

Exercise 0.18: Prove the cases $n=3$ and $n=4$ of Fermat's last theorem.

I would like to know:

  1. Is my proof correct?
  2. Is there a simpler way to prove the result using the Gaussian integers? (I'm aware of the proof that uses infinite descent, over the integers, so I would like to know if there is a very short proof using the power of the Gaussian integers)

Here is my attempt:

Let $$x^4+y^4=z^4$$ with $x,y,z\in\Bbb N$. We can assume that $x,y,z$ has no common factor, otherwise we could divide through by that factor. If $z$ is even, then $x^4+y^4\equiv_4 0$, which is only possible if $x^4,y^4\equiv_4 0$, but then $x,y,z$ all have a common factor. So assume that $x$ is even. Write $$(x^2+iy^2)(x^2-iy^2)=z^4$$

Let $\pi\in\Bbb Z[i]$ be a prime such that $\pi\mid x^2+iy^2, x^2-iy^2$. Then, $\pi\mid 2x^2,2y^2$. If $\pi=1-i$ then $N(\pi)=2\mid z$. Contradiction.

Therefore $x^2+iy^2,x^2-iy^2$ are relatively prime. We can write $x^2+iy^2=i^k\alpha^4=i^k(a+bi)^4$, and we get $$x^2+y^2i=i^k(a^4-6a^2b^2+b^4+i4ab(a^2-b^2))$$ $k=0,2$ is impossible, since this implies that $y$ is even, so $k=1,3$.

Therefore $$\begin{align}x^2&=\pm 4ab(a^2-b^2)\\y^2&=\mp (a^4-6a^2b^2+b^4)\end{align}$$ Since $y$ is odd, we must have that $a,b$ have different parities. If $k=3$ we get a contradiction no matter which of $a,b$ we assume to be even, since then $y^2\equiv_4 -a^4$ or $y^2\equiv_4 -b^4$, which is impossible. Therefore $k=1$, and so $$\begin{align}x^2&=4ab(b^2-a^2)\\y^2&=a^4-6a^2b^2+b^4=(b^2-2ab-a^2)(b^2+2ab-a^2)\end{align}$$ The factors of $b^2-2ab-a^2,b^2+2ab-a^2$ are relatively prime so they must both be odd squares. To see this, let $d\mid b^2-2ab-a^2,b^2+2ab-a^2$. Then $d\mid b^2-a^2, ab$, in particular $d\mid a^2b^2-a^2,b^4-a^2b^2$.

Let therefore $$C^2=b^2-2ab-a^2$$ solving for $b$ yields $$b=1\pm\sqrt{2+\frac{C^2}{a^2}}$$ which implies that $a\mid C\Rightarrow a\mid y$. Hence both $x,y$ have an even factor of $a$. This is only possible if $a=0$.

cansomeonehelpmeout
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  • You haven't really justified why $z$ cannot be even. – user10354138 Apr 25 '21 at 17:13
  • @user10354138 The squares modulo $4$ are $0,1$. If $z$ is even, then $z^4\equiv_4 0$, and so $x^4+y^4\equiv_4 0$, this is only possible if both $x,y$ are even, but then $x,y,z$ have a common factor. Thank you, I'll update my post! – cansomeonehelpmeout Apr 25 '21 at 17:16
  • @user10354138 If $z$ is even, then $x$ and $y$ must have the same parity. They can't both be even because if they were, $2$ would be a common factor of $x, y,$ and $z$. They can't both be odd for the reason in text, since $x$ odd $\Rightarrow x^4 \equiv 1 \pmod 4$. – Robert Shore Apr 25 '21 at 17:19
  • Proof looks good to me. One simplification is that primes by definition are not units. – Eric Apr 25 '21 at 17:41
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    @DietrichBurde "I'm doing the second part of the following exercise in Miles Reid's Undergraduate Commutative Algebra". The first part is a seperate question – cansomeonehelpmeout Apr 25 '21 at 18:31
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    I would be both satisfied and grateful if the downvoter explained how this post could be improved. – cansomeonehelpmeout Apr 26 '21 at 07:16

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