[proofplan] The proof uses the elementary classification of integer squares modulo $4$: an even square is congruent to $0$, and an odd square is congruent to $1$. This immediately rules out an odd-odd representation of a square as $a^2+b^2$, because the left-hand side is congruent to $2$ modulo $4$. For the equation $a^2+b^2=3c^2$, reducing modulo $4$ first forces $c$ to be even, and then the same square-residue classification forces $a$ and $b$ to be even. [/proofplan] [step:Classify integer squares modulo $4$] Let $n \in \mathbb{Z}$. If $n$ is even, then there exists $k \in \mathbb{Z}$ such that $n = 2k$, and hence \begin{align*} n^2 = 4k^2 \equiv 0 \pmod{4}. \end{align*} If $n$ is odd, then there exists $k \in \mathbb{Z}$ such that $n = 2k+1$, and hence \begin{align*} n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1)+1 \equiv 1 \pmod{4}. \end{align*} Therefore every integer square is congruent to either $0$ or $1$ modulo $4$, with odd squares congruent to $1$ modulo $4$ and even squares congruent to $0$ modulo $4$. [/step] [step:Rule out a square as the sum of two odd squares] Suppose, for contradiction, that $a,b,c \in \mathbb{Z}$, that $a$ and $b$ are both odd, and that \begin{align*} a^2+b^2=c^2. \end{align*} By the square classification modulo $4$, since $a$ and $b$ are odd, \begin{align*} a^2+b^2 \equiv 1+1 \equiv 2 \pmod{4}. \end{align*} On the other hand, applying the same classification to $c$, we have \begin{align*} c^2 \equiv 0 \pmod{4} \qquad \text{or} \qquad c^2 \equiv 1 \pmod{4}. \end{align*} Thus $c^2$ cannot be congruent to $2$ modulo $4$, contradicting $a^2+b^2=c^2$. Therefore no such integers $a,b,c$ exist. [/step] [step:Force $c$ to be even in the equation $a^2+b^2=3c^2$] Let $a,b,c \in \mathbb{Z}$ satisfy \begin{align*} a^2+b^2=3c^2. \end{align*} Assume, for contradiction, that $c$ is odd. Then the square classification gives $c^2 \equiv 1 \pmod{4}$, so \begin{align*} 3c^2 \equiv 3 \pmod{4}. \end{align*} Meanwhile $a^2$ and $b^2$ are each congruent to either $0$ or $1$ modulo $4$, so \begin{align*} a^2+b^2 \equiv 0,1,\text{ or }2 \pmod{4}. \end{align*} This contradicts $a^2+b^2=3c^2$. Hence $c$ is even. [/step] [step:Force $a$ and $b$ to be even once $c$ is even] Since $c$ is even, the square classification gives \begin{align*} c^2 \equiv 0 \pmod{4}. \end{align*} Using $a^2+b^2=3c^2$, we obtain \begin{align*} a^2+b^2 \equiv 0 \pmod{4}. \end{align*} Each of $a^2$ and $b^2$ is congruent to either $0$ or $1$ modulo $4$. The only way their sum is congruent to $0$ modulo $4$ is that both are congruent to $0$ modulo $4$. Therefore \begin{align*} a^2 \equiv 0 \pmod{4} \qquad \text{and} \qquad b^2 \equiv 0 \pmod{4}. \end{align*} By the square classification, this means $a$ and $b$ are even. Together with the previous step, $a$, $b$, and $c$ are all even. [/step]