[proofplan] We first determine which of the two coprime factors contains the unique parity contribution from the factor $2$. After possibly interchanging $a$ and $b$, we may assume that $b$ is even and $a$ is odd. Prime-exponent parity in the identity $ab=2c^2$ then forces every exponent in $a$ to be even, while the exponent of $2$ in $b$ is odd and every odd-prime exponent in $b$ is even. This gives $a=r^2$ and $b=2s^2$, and coprimality of $r$ and $s$ follows from coprimality of $a$ and $b$. [/proofplan] [step:Interchange the factors so that the even factor is $b$] Since $ab=2c^2$, the product $ab$ is even. Hence at least one of $a$ and $b$ is even. Since $\gcd(a,b)=1$, they cannot both be even. Therefore exactly one of $a$ and $b$ is even. If necessary, interchange the names of $a$ and $b$ so that $b$ is even. Then $a$ is odd and $\gcd(a,b)=1$ still holds. [/step] [step:Use prime-exponent parity to show that the odd factor is a square] For a prime number $p$ and an integer $n \in \mathbb{N}$, let $\nu_p(n) \in \mathbb{N}\cup\{0\}$ denote the exponent of $p$ in the prime factorization of $n$. Since $a$ is odd, $\nu_2(a)=0$. Let $p$ be an odd prime. From $\gcd(a,b)=1$, at most one of $\nu_p(a)$ and $\nu_p(b)$ is nonzero. The identity $ab=2c^2$ gives \begin{align*} \nu_p(a)+\nu_p(b)=\nu_p(ab)=\nu_p(2c^2)=2\nu_p(c), \end{align*} because $p$ is odd and therefore $\nu_p(2)=0$. Thus $\nu_p(a)+\nu_p(b)$ is even. If $\nu_p(a)>0$, then $\nu_p(b)=0$, so \begin{align*} \nu_p(a)=2\nu_p(c). \end{align*} Hence every prime exponent in $a$ is even. Define $r \in \mathbb{N}$ by \begin{align*} r=\prod_{p\ \mathrm{prime}} p^{\nu_p(a)/2}, \end{align*} where only finitely many factors are different from $1$. Then \begin{align*} r^2=\prod_{p\ \mathrm{prime}} p^{\nu_p(a)}=a. \end{align*} So $a=r^2$. [/step] [step:Use the remaining parity to write the even factor as twice a square] We now analyze the prime exponents of $b$. For the prime $2$, the identity $ab=2c^2$ and $\nu_2(a)=0$ give \begin{align*} \nu_2(b)=\nu_2(ab)=\nu_2(2c^2)=1+2\nu_2(c). \end{align*} Thus $\nu_2(b)-1$ is even. For an odd prime $p$, if $\nu_p(b)>0$, then $\nu_p(a)=0$ because $\gcd(a,b)=1$. As above, \begin{align*} \nu_p(b)=\nu_p(a)+\nu_p(b)=\nu_p(ab)=2\nu_p(c), \end{align*} so every odd-prime exponent in $b$ is even. Define $s \in \mathbb{N}$ by \begin{align*} s=2^{(\nu_2(b)-1)/2}\prod_{\substack{p\ \mathrm{prime}\\ p\ \mathrm{odd}}} p^{\nu_p(b)/2}, \end{align*} again with only finitely many nontrivial factors. Then \begin{align*} 2s^2 =2^{1+2\cdot(\nu_2(b)-1)/2}\prod_{\substack{p\ \mathrm{prime}\\ p\ \mathrm{odd}}}p^{\nu_p(b)} =2^{\nu_2(b)}\prod_{\substack{p\ \mathrm{prime}\\ p\ \mathrm{odd}}}p^{\nu_p(b)} =b. \end{align*} Therefore $b=2s^2$. [/step] [step:Deduce that the square roots are coprime] It remains to prove $\gcd(r,s)=1$. Suppose that a prime $p$ divides both $r$ and $s$. Then $p$ divides $r^2=a$ and also divides $s^2$, hence divides $2s^2=b$. Therefore $p$ divides both $a$ and $b$, contradicting $\gcd(a,b)=1$. Thus no prime divides both $r$ and $s$, so $\gcd(r,s)=1$. Combining this with the previous steps gives, after the possible initial interchange of $a$ and $b$, \begin{align*} a=r^2,\qquad b=2s^2, \end{align*} with $\gcd(r,s)=1$. [/step]