[proofplan] We prove the theorem by infinite descent. Assuming a counterexample exists, we choose one with minimal hypotenuse and first reduce to a primitive Pythagorean triple. The primitive triple parametrization converts the square-area condition into a product of pairwise coprime integers being a square, forcing those integers themselves to be squares. A second parametrization then constructs a new square-area right triangle with strictly smaller hypotenuse, contradicting the minimal choice. [/proofplan]

[step:Choose a primitive counterexample with minimal hypotenuse] Assume, for contradiction, that there exist positive integers $a,b,c,N \in \mathbb{N}$ such that \begin{align*} a^2+b^2=c^2, \qquad \frac{ab}{2}=N^2. \end{align*} Among all such quadruples, choose one with $c$ minimal. Let $d=\gcd(a,b)$. Since $a^2+b^2=c^2$, we have $d \mid c$. Define positive integers \begin{align*} x := \frac{a}{d}, \qquad y := \frac{b}{d}, \qquad z := \frac{c}{d}. \end{align*} Then $\gcd(x,y)=1$ and \begin{align*} x^2+y^2=z^2. \end{align*} Moreover, \begin{align*} \frac{xy}{2} = \frac{ab}{2d^2} = \frac{N^2}{d^2} = \left(\frac{N}{d}\right)^2. \end{align*} Since $d \mid a$ and $d \mid b$, the equality $ab=2N^2$ implies $d^2 \mid N^2$, hence $d \mid N$ by prime factorization. Thus $N/d \in \mathbb{N}$, so $(x,y,z)$ is again a square-area right triangle. Minimality of $c$ gives $d=1$. Therefore the chosen counterexample is primitive: $\gcd(a,b)=1$. Rename this primitive counterexample $(x,y,z,N)$, so \begin{align*} \gcd(x,y)=1, \qquad x^2+y^2=z^2, \qquad \frac{xy}{2}=N^2, \end{align*} and $z$ is minimal among all square-area right triangles. [/step]

[step:Parametrize the primitive Pythagorean triple] Because $\gcd(x,y)=1$ and $x^2+y^2=z^2$, exactly one of $x,y$ is even. Interchanging $x$ and $y$ if necessary, assume $x$ is odd and $y$ is even. [claim:Primitive Pythagorean triples have Euclid parameters] There exist positive integers $m,n \in \mathbb{N}$ such that \begin{align*} m>n, \qquad \gcd(m,n)=1, \qquad m \not\equiv n \pmod 2, \end{align*} and \begin{align*} x=m^2-n^2, \qquad y=2mn, \qquad z=m^2+n^2. \end{align*} [/claim] [proof] Since $x$ is odd, both $z-x$ and $z+x$ are even. Define positive integers \begin{align*} u := \frac{z+x}{2}, \qquad v := \frac{z-x}{2}. \end{align*} Then \begin{align*} uv = \frac{(z+x)(z-x)}{4} = \frac{z^2-x^2}{4} = \frac{y^2}{4} = \left(\frac{y}{2}\right)^2. \end{align*} Also $\gcd(u,v)=1$. Indeed, if a prime $p$ divides both $u$ and $v$, then $p$ divides $u+v=z$ and $u-v=x$, contradicting $\gcd(x,z)=1$, which follows from $\gcd(x,y)=1$ and $x^2+y^2=z^2$. Since $u$ and $v$ are coprime and their product is a square, prime factorization gives positive integers $m,n \in \mathbb{N}$ with \begin{align*} u=m^2, \qquad v=n^2. \end{align*} Because $u>v$, we have $m>n$. Hence \begin{align*} z+x=2m^2, \qquad z-x=2n^2. \end{align*} Adding and subtracting these equations gives \begin{align*} z=m^2+n^2, \qquad x=m^2-n^2. \end{align*} Finally, \begin{align*} y^2=z^2-x^2=(z-x)(z+x)=4m^2n^2, \end{align*} so $y=2mn$. If a prime $p$ divides both $m$ and $n$, then $p^2$ divides both $u$ and $v$, contradicting $\gcd(u,v)=1$. Thus $\gcd(m,n)=1$. If $m$ and $n$ had the same parity, then both would be odd, and $x=m^2-n^2$ would be even, contradicting that $x$ is odd. Therefore $m$ and $n$ have opposite parity. [/proof] [/step]

[step:Use the square-area condition to force three coprime factors to be squares] The area condition and the parametrization give \begin{align*} N^2 = \frac{xy}{2} = \frac{(m^2-n^2)(2mn)}{2} = mn(m^2-n^2). \end{align*} The three positive integers $m$, $n$, and $m^2-n^2$ are pairwise coprime. Indeed, \begin{align*} \gcd(m,n)=1, \end{align*} and \begin{align*} \gcd(m,m^2-n^2)=\gcd(m,n^2)=1, \qquad \gcd(n,m^2-n^2)=\gcd(n,m^2)=1. \end{align*} Since their product is a square and they are pairwise coprime, prime factorization implies that each factor is a square. Hence there exist positive integers $r,s,t \in \mathbb{N}$ such that \begin{align*} m^2-n^2=r^2, \qquad m=s^2, \qquad n=t^2. \end{align*} Consequently, \begin{align*} r^2+t^4=s^4. \end{align*} Thus $(r,t^2,s^2)$ is a primitive Pythagorean triple, because \begin{align*} r^2+(t^2)^2=(s^2)^2. \end{align*} It is primitive since any common divisor of $r$ and $t^2$ would divide $m^2-n^2$ and $n$, hence would divide $m$, contradicting $\gcd(m,n)=1$. [/step]

[step:Parametrize the smaller auxiliary triple] The triple $(r,t^2,s^2)$ is primitive. Since $m$ and $n$ have opposite parity, $s$ and $t$ have opposite parity. Therefore $t^2$ is even and $r$ is odd after the usual choice of the odd leg in the primitive parametrization. Applying the parametrization proved above to this primitive triple gives positive integers $p,q \in \mathbb{N}$ such that \begin{align*} p>q, \qquad \gcd(p,q)=1, \qquad p \not\equiv q \pmod 2, \end{align*} and \begin{align*} r=p^2-q^2, \qquad t^2=2pq, \qquad s^2=p^2+q^2. \end{align*} Because $p$ and $q$ are coprime and $2pq=t^2$ is a square, exactly one of $p,q$ is even, and prime factorization gives positive integers $u,v \in \mathbb{N}$ such that either \begin{align*} p=u^2, \qquad q=2v^2, \end{align*} or \begin{align*} p=2v^2, \qquad q=u^2. \end{align*} In both cases, \begin{align*} s^2=u^4+4v^4. \end{align*} Equivalently, \begin{align*} (u^2)^2+(2v^2)^2=s^2. \end{align*} [/step]

[step:Construct a smaller square-area right triangle] The positive integers \begin{align*} X := u^2, \qquad Y := 2v^2, \qquad Z := s \end{align*} satisfy \begin{align*} X^2+Y^2 = (u^2)^2+(2v^2)^2 = s^2 = Z^2. \end{align*} Their area is \begin{align*} \frac{XY}{2} = \frac{u^2\cdot 2v^2}{2} = (uv)^2, \end{align*} so $(X,Y,Z)$ is another square-area right triangle. It remains to compare hypotenuses. The original primitive counterexample has \begin{align*} z=m^2+n^2=s^4+t^4. \end{align*} Since $s,t \in \mathbb{N}$, we have \begin{align*} s < s^4+t^4=z. \end{align*} Thus the newly constructed square-area right triangle has hypotenuse $Z=s$ strictly smaller than the minimal hypotenuse $z$, a contradiction. [/step]

[step:Conclude that no square-area integer right triangle exists] The contradiction shows that the assumed counterexample cannot exist. Therefore there are no positive integers $x,y,z,N \in \mathbb{N}$ satisfying \begin{align*} x^2+y^2=z^2, \qquad \frac{xy}{2}=N^2. \end{align*} This proves that no right triangle with integer side lengths has square area. [/step]