[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]
