[proofplan] We argue by infinite descent. Assuming a solution exists, choose one with minimal $z$ and first reduce it to a primitive solution. The equation makes $(x^2,y^2,z)$ a primitive Pythagorean triple, so a complete parametrisation expresses the two legs as $m^2-n^2$ and $2mn$. Square-splitting in coprime products then forces a second square decomposition, from which we construct a new positive solution $X^4+Y^4=Z^2$ with $Z<z$, contradicting the minimal choice of $z$. [/proofplan] [step:Choose a primitive counterexample with minimal hypotenuse] Assume, for contradiction, that there exist positive integers $x,y,z \in \mathbb{N}$ satisfying \begin{align*} x^4+y^4=z^2. \end{align*} Among all such triples, choose one for which $z$ is minimal. We first show that $\gcd(x,y)=1$. Let $d:=\gcd(x,y)$, and write $x=d x_1$ and $y=d y_1$ with $x_1,y_1 \in \mathbb{N}$ and $\gcd(x_1,y_1)=1$. Then \begin{align*} d^4(x_1^4+y_1^4)=z^2. \end{align*} Thus $d^4$ divides $z^2$, so $d^2$ divides $z$. Hence there is $z_1 \in \mathbb{N}$ such that $z=d^2z_1$, and division by $d^4$ gives \begin{align*} x_1^4+y_1^4=z_1^2. \end{align*} If $d>1$, then $z_1<z$, contradicting the minimality of $z$. Therefore $d=1$, so $\gcd(x,y)=1$. Since $\gcd(x,y)=1$, the integers $x$ and $y$ are not both even. They are also not both odd, because fourth powers of odd integers are congruent to $1$ modulo $16$, so $x^4+y^4 \equiv 2 \pmod{16}$, while a square modulo $16$ is congruent to $0$, $1$, $4$, or $9$. Therefore exactly one of $x,y$ is even. Interchanging $x$ and $y$ if necessary, assume from now on that $y$ is even and $x$ is odd. Then \begin{align*} (x^2)^2+(y^2)^2=z^2, \end{align*} and $\gcd(x^2,y^2)=1$, so $(x^2,y^2,z)$ is a primitive Pythagorean triple with even leg $y^2$. [/step] [step:Parametrize the primitive Pythagorean triple] We use the following elementary form of the primitive Pythagorean triple parametrisation. [claim:Primitive Pythagorean triples have Euclidean parameters] Let $A,B,C \in \mathbb{N}$ satisfy \begin{align*} A^2+B^2=C^2, \end{align*} with $\gcd(A,B)=1$ and $B$ even. Then there exist coprime positive integers $m,n \in \mathbb{N}$ with $m>n$ and opposite parity such that \begin{align*} A=m^2-n^2,\qquad B=2mn,\qquad C=m^2+n^2. \end{align*} [/claim] [proof] Since $\gcd(A,B)=1$ and $B$ is even, $A$ is odd. Hence $C^2=A^2+B^2$ is odd, so $C$ is odd. Define positive integers \begin{align*} R:=\frac{C+A}{2},\qquad S:=\frac{C-A}{2}. \end{align*} Then $R>S>0$ and \begin{align*} RS=\frac{(C+A)(C-A)}{4}=\frac{C^2-A^2}{4}=\frac{B^2}{4}. \end{align*} We claim that $\gcd(R,S)=1$. If a positive integer $q$ divides both $R$ and $S$, then $q$ divides $R+S=C$ and $R-S=A$. Hence $q$ divides $\gcd(A,C)$. But any common divisor of $A$ and $C$ also divides $C^2-A^2=B^2$, and therefore divides $B$; since $\gcd(A,B)=1$, we get $q=1$. Since $R$ and $S$ are coprime and their product is the square $(B/2)^2$, unique prime factorisation implies that each is a square. Thus there exist positive integers $m,n \in \mathbb{N}$ such that \begin{align*} R=m^2,\qquad S=n^2. \end{align*} Then \begin{align*} A=R-S=m^2-n^2,\qquad C=R+S=m^2+n^2,\qquad \frac{B^2}{4}=m^2n^2. \end{align*} Since $B,m,n$ are positive, the last equality gives $B=2mn$. The relation $\gcd(R,S)=1$ gives $\gcd(m,n)=1$. Finally, $m$ and $n$ have opposite parity, because if they were both odd then $A=m^2-n^2$ would be even, contradicting that $A$ is odd; they cannot both be even because they are coprime. [/proof] Applying the claim with $A=x^2$, $B=y^2$, and $C=z$, we obtain coprime positive integers $m,n \in \mathbb{N}$ with $m>n$ and opposite parity such that \begin{align*} x^2=m^2-n^2,\qquad y^2=2mn,\qquad z=m^2+n^2. \end{align*} [/step] [step:Split the coprime product $2mn$ into a square and twice a square] Because $\gcd(m,n)=1$ and \begin{align*} 2mn=y^2, \end{align*} unique prime factorisation implies that the odd member of $\{m,n\}$ is a square and the even member is twice a square. Indeed, every odd prime exponent in $2mn$ is even; since $m$ and $n$ are coprime, the odd prime exponents in each of $m$ and $n$ are separately even. The single extra factor of $2$ must occur in the even member, and its exponent there is odd, so that even member is twice a square. We next exclude the possibility that $m$ is even. If $m$ were even and $n$ odd, then there would be positive integers $r,s \in \mathbb{N}$ such that \begin{align*} m=2r^2,\qquad n=s^2. \end{align*} Then \begin{align*} x^2=m^2-n^2=(m-n)(m+n). \end{align*} The integers $m-n$ and $m+n$ are odd. Their greatest common divisor divides both their sum $2m$ and their difference $2n$; since it is odd and $\gcd(m,n)=1$, their greatest common divisor is $1$. Hence each of the coprime factors $m-n$ and $m+n$ is a square. Thus there exist odd positive integers $u,v \in \mathbb{N}$ with $v>u$ such that \begin{align*} m-n=u^2,\qquad m+n=v^2. \end{align*} Subtracting gives \begin{align*} v^2-u^2=2n=2s^2. \end{align*} But odd squares are congruent to $1$ modulo $8$, so $v^2-u^2 \equiv 0 \pmod{8}$, whereas $2s^2 \equiv 2 \pmod{8}$ because $s$ is odd. This contradiction shows that $m$ is not even. Therefore $m$ is odd and $n$ is even. Hence there exist positive integers $a,b \in \mathbb{N}$ such that \begin{align*} m=a^2,\qquad n=2b^2. \end{align*} [/step] [step:Split $m^2-n^2$ into two coprime squares] From \begin{align*} x^2=m^2-n^2=(m-n)(m+n), \end{align*} we show that both factors are squares. Since $m$ is odd and $n$ is even, the integers $m-n$ and $m+n$ are odd. If $q$ is a common divisor of $m-n$ and $m+n$, then $q$ divides their sum $2m$ and their difference $2n$. Because $q$ is odd and $\gcd(m,n)=1$, this forces $q=1$. Thus \begin{align*} \gcd(m-n,m+n)=1. \end{align*} Since the product $(m-n)(m+n)$ is the square $x^2$ and the two factors are coprime, unique prime factorisation implies that each factor is a square. Therefore there exist odd positive integers $u,v \in \mathbb{N}$ with $v>u$ such that \begin{align*} m-n=u^2,\qquad m+n=v^2. \end{align*} Subtracting and using $n=2b^2$ gives \begin{align*} v^2-u^2=2n=4b^2. \end{align*} Since $u$ and $v$ are odd, define positive integers \begin{align*} P:=\frac{v-u}{2},\qquad Q:=\frac{v+u}{2}. \end{align*} Then \begin{align*} PQ=\frac{v^2-u^2}{4}=b^2. \end{align*} Also $\gcd(P,Q)=1$: any common divisor of $P$ and $Q$ divides $P+Q=v$ and $Q-P=u$, and $\gcd(u,v)=1$ because $u^2=m-n$ and $v^2=m+n$ are coprime. Hence $P$ and $Q$ are coprime positive integers whose product is a square, so each is a square. Thus there exist positive integers $X,Y \in \mathbb{N}$ such that \begin{align*} P=X^2,\qquad Q=Y^2. \end{align*} Because $Q>P$, we have $Y>X$. [/step] [step:Construct a smaller fourth-power solution] From the definitions of $P$ and $Q$, we have \begin{align*} u=Q-P=Y^2-X^2,\qquad v=Q+P=Y^2+X^2. \end{align*} Using \begin{align*} m-n=u^2,\qquad m+n=v^2, \end{align*} we compute \begin{align*} 2m=u^2+v^2 &=(Y^2-X^2)^2+(Y^2+X^2)^2 \\ &=2X^4+2Y^4. \end{align*} Dividing by $2$ gives \begin{align*} m=X^4+Y^4. \end{align*} But $m=a^2$, so \begin{align*} X^4+Y^4=a^2. \end{align*} Thus $(X,Y,a)$ is another positive integer solution of the same equation. It remains to compare the new square root $a$ with the original minimal square root $z$. Since $m=a^2$ and $n>0$, we have \begin{align*} a^2=m<m^2+n^2=z. \end{align*} Because $a,z \in \mathbb{N}$ and $z\geq 1$, the inequality $a^2<z$ implies $a<z$. Hence $(X,Y,a)$ is a positive solution with smaller third component than the chosen minimal solution $(x,y,z)$, a contradiction. Therefore no positive integers $x,y,z \in \mathbb{N}$ satisfy \begin{align*} x^4+y^4=z^2. \end{align*} [/step]