Show that there are no positive integer solutions to $x^4-y^4=z^4$

Question

Also, Show that the only solution in relatively prime positive integers to $x^4+y^4=2z^4$ is $x=y=z=1$.

Here I tried a similar contradiction proof to the proof for $x^4+y^4=z^4$, but this doesn't work the same way.

Answer 1

We show that if there exist $x,y,z \in \mathbb N$ with $x^2=y^4+z^4$ then there exist $x',y',z'\in \mathbb N$ with $x'^2=y'^4+z'^4$ and $x'<x.$ This implies that, if $S=\{x\in \mathbb N:\;\exists y,z \in \mathbb N\;(x^2=y^4+z^4)\}$ is not empty, then $S$ is a non-empty subset of $\mathbb N$ with no least member, which is not possible. So we will conclude that $S$ is empty.

Assume $x^2=y^4+z^4.$

(i). If $\gcd(y,z)>1,$ let $p$ be a prime divisor of $\gcd (y,z).$ Then $p^4 $ divides $x^2.$ Since $p$ is prime this implies that $p^2$ divides $x.$ So let $x'=x/p^2,\;y'=y/p,\;z'=z/p.$

(ii). If $\gcd (y,z)=1$ then $(y^2,z^2,x)$ is a primitive Pythagorean triplet. So there exist $m,n\in \mathbb N,$ not both odd, with $$\{y^2,z^2\}=\{m^2-n^2,2mn\} \;\land \;x=m^2+n^2$$and with $\gcd(m,n)=1.$... Now WLOG, $$y^2=m^2-n^2 \;\land \;z^2=2mn.$$

Note that $y$ is odd. Else $2$ divides both $y^2$ and $z^2=2mn,$ implying that $2$ divides both $y$ and $z,$ contrary to $\gcd (y,z)=1.$

Now $y^2+n^2=m^2$ and we must have $\gcd (y,n)=1.$ Because if a prime $p$ divides both $y$ and $n$ then $p^2$ divides $m^2=y^2+n^2,$ implying (since $p$ is prime) that $p$ divides $m ,$ implying $\gcd(m,n)\geq p>1,$ contrary to $\gcd (m,n)=1.$

Therefore $(y,n,m)$ is a primitive Pythagorean triplet. So there exist $a,b \in \mathbb N,$ not both odd, with $$\{y,n\}=\{a^2-b^2,2ab\} \;\land\; m=a^2+b^2$$ and with $\gcd(a,b)=1.$ Now since $y$ is odd we have $$y=a^2-b^2 \;\land \;n=2ab.$$

Applying this to $z,$ we have $$z^2=2mn=2(a^2+b^2)(2ab)=4(a^2+b^2)ab.$$ So $z/2\in \mathbb N,$ and $(a^2+b^2)ab=(z/2)^2$ is the square of a member of $\mathbb N.$ Now any prime that divides any two of $(a^2+b^2,a,b)$ will divide the third one. But $\gcd (a,b)=1.$ So $(a^2+b^2,a,b)$ are pair-wise co-prime.

From the uniqueness of prime decompositions on $\mathbb N$ we have: For any pair-wise co-prine $(U,V,W)$ in $\mathbb N$, if $UVW$ is the square of a member of $\mathbb N$ then each of $U,V,W$ is the square of a member of $\mathbb N.$ Applying this with $U=a^2+b^2,\; V=a,\;W=b$ we infer that there exist $x',y',z' \in \mathbb N$ with $$a^2+b^2=x'^2\;\land\; a=y'^2\;\land\; b=z'^2.$$ Therefore $$x'^2=a^2+b^2=y'^4+z'^4.$$ And we have $x'=\sqrt {a^2+b^2}\;=\sqrt m<2mn=z^2<\sqrt {z^4+y^4}=\sqrt {x^2}=x.$