[proofplan] We first prove that in a primitive Pythagorean triple with $y$ even, both $x$ and $z$ are odd. This makes the two integers $(z+x)/2$ and $(z-x)/2$ positive, coprime, and with product $(y/2)^2$. A coprime square-factor lemma then forces these two integers to be squares, giving the parameters $m$ and $n$ and the formulas. Finally, we prove uniqueness from the identities for $(z+x)/2$ and $(z-x)/2$, and verify directly that every admissible pair $(m,n)$ gives a primitive Pythagorean triple. [/proofplan] [step:Show that $x$ and $z$ are odd] Since $y$ is even, $y^2$ is divisible by $4$. From $x^2+y^2=z^2$, we have \begin{align*} x^2 \equiv z^2 \pmod 4. \end{align*} If $x$ were even, then $z$ would also be even, and then $2$ would divide $x,y,z$, contradicting $\gcd(x,y,z)=1$. Hence $x$ is odd. Since $x^2 \equiv z^2 \pmod 4$ and an odd square is congruent to $1$ modulo $4$, it follows that $z$ is odd. [guided] The parity hypothesis is used first. Because $y$ is even, there is an integer $k \in \mathbb N$ with $y=2k$, so \begin{align*} y^2 = 4k^2. \end{align*} Thus $y^2$ is divisible by $4$. Reducing the Pythagorean equation modulo $4$ gives \begin{align*} x^2+y^2=z^2 \quad\Longrightarrow\quad x^2 \equiv z^2 \pmod 4. \end{align*} Now an integer square is congruent to either $0$ or $1$ modulo $4$: it is congruent to $0$ if the integer is even and to $1$ if the integer is odd. If $x$ were even, then $x^2 \equiv 0 \pmod 4$, so $z^2 \equiv 0 \pmod 4$, forcing $z$ to be even. But then $2$ would divide all of $x,y,z$, contradicting the primitive hypothesis $\gcd(x,y,z)=1$. Therefore $x$ is odd. Since $x$ is odd, $x^2 \equiv 1 \pmod 4$. The congruence $x^2 \equiv z^2 \pmod 4$ gives $z^2 \equiv 1 \pmod 4$, so $z$ is odd. [/guided] [/step] [step:Construct two positive coprime factors whose product is a square] Define integers $A,B \in \mathbb N$ by \begin{align*} A := \frac{z+x}{2}, \qquad B := \frac{z-x}{2}. \end{align*} These are integers because $x$ and $z$ are odd. Since $z^2=x^2+y^2>x^2$, we have $z>x$, so $A>B>0$. Moreover, \begin{align*} AB = \frac{(z+x)(z-x)}{4} = \frac{z^2-x^2}{4} = \frac{y^2}{4} = \left(\frac{y}{2}\right)^2. \end{align*} We claim that $\gcd(A,B)=1$. Let $d \in \mathbb N$ divide both $A$ and $B$. Then $d$ divides \begin{align*} A+B=z \end{align*} and \begin{align*} A-B=x. \end{align*} Thus $d$ divides both $x$ and $z$. Since $d$ also divides $A+B$ and $A-B$, any odd prime divisor of $d$ would divide both $x$ and $z$, hence also $y^2=z^2-x^2$, and therefore would divide $y$, contradicting $\gcd(x,y,z)=1$. The only possible common divisor of $A$ and $B$ is therefore a power of $2$. But $A+B=z$ is odd, so $A$ and $B$ cannot both be even. Hence $\gcd(A,B)=1$. [guided] The classical parametrisation comes from factoring the difference of squares: \begin{align*} z^2-x^2=y^2. \end{align*} Because $x$ and $z$ are odd, the half-sum and half-difference are integers. Define \begin{align*} A := \frac{z+x}{2}, \qquad B := \frac{z-x}{2}. \end{align*} The inequality $z>x$ follows from $z^2=x^2+y^2$ and $y>0$, so $A$ and $B$ are positive integers with $A>B$. Their product is \begin{align*} AB = \frac{(z+x)(z-x)}{4} = \frac{z^2-x^2}{4} = \frac{y^2}{4} = \left(\frac{y}{2}\right)^2. \end{align*} Thus $A$ and $B$ are two positive integer factors of a square. We next prove that these two factors are coprime. Let $d \in \mathbb N$ be a common divisor of $A$ and $B$. Then $d$ divides their sum and difference: \begin{align*} A+B=z, \qquad A-B=x. \end{align*} So every prime divisor of $d$ divides both $x$ and $z$. If an odd prime $p$ divides both $x$ and $z$, then $p$ divides $z^2-x^2=y^2$, so $p$ divides $y$. This contradicts $\gcd(x,y,z)=1$. Hence no odd prime divides $d$. It remains to exclude $2$. If $2$ divided both $A$ and $B$, then $2$ would divide $A+B=z$, contradicting that $z$ is odd. Therefore no prime divides both $A$ and $B$, and hence \begin{align*} \gcd(A,B)=1. \end{align*} [/guided] [/step] [step:Use the coprime square factor lemma to obtain $m$ and $n$] [claim:Coprime factors of a square are squares] Let $A,B \in \mathbb N$ satisfy $\gcd(A,B)=1$ and suppose $AB$ is a square. Then there exist $r,s \in \mathbb N$ such that $A=r^2$ and $B=s^2$. [/claim] [proof] Write the prime factorizations \begin{align*} A=\prod_{i=1}^{k} p_i^{\alpha_i}, \qquad B=\prod_{j=1}^{\ell} q_j^{\beta_j}, \end{align*} where the $p_i$ and $q_j$ are primes and all exponents $\alpha_i,\beta_j$ are positive integers. Since $\gcd(A,B)=1$, no prime $p_i$ equals any prime $q_j$. Therefore the prime factorization of $AB$ is obtained by combining these two disjoint lists. Since $AB$ is a square, every prime exponent in the factorization of $AB$ is even. Hence every $\alpha_i$ and every $\beta_j$ is even. Define \begin{align*} r := \prod_{i=1}^{k} p_i^{\alpha_i/2}, \qquad s := \prod_{j=1}^{\ell} q_j^{\beta_j/2}. \end{align*} Then $A=r^2$ and $B=s^2$. [/proof] Apply the claim to the coprime positive integers $A$ and $B$. Since \begin{align*} AB=\left(\frac{y}{2}\right)^2, \end{align*} there exist $m,n \in \mathbb N$ such that \begin{align*} A=m^2, \qquad B=n^2. \end{align*} Because $A>B$, we have $m>n$. Substituting the definitions of $A$ and $B$ gives \begin{align*} \frac{z+x}{2}=m^2, \qquad \frac{z-x}{2}=n^2. \end{align*} Adding and subtracting these equations yields \begin{align*} z=m^2+n^2, \qquad x=m^2-n^2. \end{align*} Finally, \begin{align*} \left(\frac{y}{2}\right)^2=AB=m^2n^2. \end{align*} Since $y,m,n$ are positive, this gives \begin{align*} y=2mn. \end{align*} [/step] [step:Prove that the parameters are coprime and of opposite parity] Since $A=m^2$ and $B=n^2$ are coprime, $\gcd(m,n)=1$. If $m$ and $n$ were both odd, then \begin{align*} m^2-n^2 \equiv 1-1 \equiv 0 \pmod 2, \qquad m^2+n^2 \equiv 1+1 \equiv 0 \pmod 2, \end{align*} so $x$ and $z$ would both be even, contradicting that $x$ and $z$ are odd. If $m$ and $n$ were both even, then $\gcd(m,n)\ge 2$, contradicting $\gcd(m,n)=1$. Hence $m$ and $n$ have opposite parity. [/step] [step:Prove uniqueness of the parameters] Suppose $m,n,m',n' \in \mathbb N$ satisfy $m>n$, $m'>n'$, and \begin{align*} x=m^2-n^2=m'^2-n'^2, \qquad z=m^2+n^2=m'^2+n'^2. \end{align*} Adding the two identities gives \begin{align*} 2m^2=x+z=2m'^2, \end{align*} so $m^2=m'^2$. Since $m,m' \in \mathbb N$, we get $m=m'$. Subtracting the identities gives \begin{align*} 2n^2=z-x=2n'^2, \end{align*} so $n^2=n'^2$, and since $n,n' \in \mathbb N$, we get $n=n'$. Therefore the parameters are unique. [/step] [step:Verify that Euclid's formulas give a primitive Pythagorean triple] Conversely, let $m,n \in \mathbb N$ satisfy $m>n$, $\gcd(m,n)=1$, and suppose $m,n$ have opposite parity. Define $x,y,z \in \mathbb N$ by \begin{align*} x := m^2-n^2, \qquad y := 2mn, \qquad z := m^2+n^2. \end{align*} Since $m>n$, $x>0$, and all three numbers are positive integers. Also $y$ is even. Direct expansion gives \begin{align*} x^2+y^2 &= (m^2-n^2)^2+(2mn)^2 \\ &= m^4-2m^2n^2+n^4+4m^2n^2 \\ &= m^4+2m^2n^2+n^4 \\ &= (m^2+n^2)^2 \\ &= z^2. \end{align*} It remains to prove primitivity. Let $d:=\gcd(x,y,z)$. Since $m$ and $n$ have opposite parity, $m^2-n^2$ and $m^2+n^2$ are odd, while $2mn$ is even. Thus $2\nmid d$. Now let $p$ be an odd prime divisor of $d$. Then $p$ divides $y=2mn$, and since $p$ is odd, $p$ divides $mn$. Hence $p$ divides $m$ or $p$ divides $n$. If $p$ divides $m$, then from $p\mid z=m^2+n^2$ we get $p\mid n^2$, so $p\mid n$, contradicting $\gcd(m,n)=1$. If $p$ divides $n$, then from $p\mid z=m^2+n^2$ we get $p\mid m^2$, so $p\mid m$, again contradicting $\gcd(m,n)=1$. Therefore no odd prime divides $d$. Since neither $2$ nor any odd prime divides $d$, we have $d=1$. Thus $(x,y,z)$ is a primitive Pythagorean triple with $y$ even. This completes both directions of the theorem. [/step]