[proofplan] We argue by contradiction from a primitive integral solution. Reducing the equation modulo $3$ shows that both $x$ and $y$ must be divisible by $3$, because the only square residues modulo $3$ are $0$ and $1$. Substituting $x=3a$ and $y=3b$ then forces $3\mid z^2$, hence $3\mid z$ by the same residue computation. Thus $3$ divides all of $x,y,z$, contradicting $\gcd(x,y,z)=1$. [/proofplan] [step:Assume a primitive solution and reduce the equation modulo $3$] Suppose, for contradiction, that there exist integers $x,y,z\in \mathbb{Z}$ with $\gcd(x,y,z)=1$ and $x^2+y^2=3z^2$. For any integer $n\in \mathbb{Z}$, its residue modulo $3$ is one of $0,1,2$. Squaring these residues gives $0^2\equiv 0 \pmod{3}$, $1^2\equiv 1 \pmod{3}$, and $2^2\equiv 1 \pmod{3}$. Thus every integer square is congruent to either $0$ or $1$ modulo $3$. Reducing the equation $x^2+y^2=3z^2$ modulo $3$ gives $x^2+y^2\equiv 0 \pmod{3}$. Since each of $x^2$ and $y^2$ is congruent to $0$ or $1$ modulo $3$, the only way their sum is congruent to $0$ modulo $3$ is that both are congruent to $0$ modulo $3$. Therefore $x^2\equiv 0 \pmod{3}$ and $y^2\equiv 0 \pmod{3}$. By the residue computation above, this implies $3\mid x$ and $3\mid y$. [guided] Assume, aiming for a contradiction, that the claimed forbidden solution exists. Thus we have integers $x,y,z\in \mathbb{Z}$ such that $\gcd(x,y,z)=1$ and $x^2+y^2=3z^2$. The useful modulus is $3$ because the right-hand side is visibly divisible by $3$. We first record exactly what squares can look like modulo $3$. Every integer $n\in \mathbb{Z}$ has one of the residues $0,1,2$ modulo $3$. Squaring those three possibilities gives $0^2\equiv 0 \pmod{3}$, $1^2\equiv 1 \pmod{3}$, and $2^2\equiv 4\equiv 1 \pmod{3}$. So an integer square is congruent only to $0$ or $1$ modulo $3$. Now reduce the equation modulo $3$. Since $3z^2$ is divisible by $3$, the equation gives $x^2+y^2\equiv 0 \pmod{3}$. There are only four residue possibilities for the pair $(x^2,y^2)$ modulo $3$: $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. Their sums are respectively $0$, $1$, $1$, and $2$ modulo $3$. Hence the congruence $x^2+y^2\equiv 0 \pmod{3}$ forces the first case: $x^2\equiv 0 \pmod{3}$ and $y^2\equiv 0 \pmod{3}$. Using the same square-residue computation, $n^2\equiv 0 \pmod{3}$ occurs exactly when $n\equiv 0 \pmod{3}$. Therefore $3\mid x$ and $3\mid y$. [/guided] [/step] [step:Substitute the divisibility of $x$ and $y$ to force $3\mid z$] Since $3\mid x$ and $3\mid y$, choose integers $a,b\in \mathbb{Z}$ such that $x=3a$ and $y=3b$. Substituting these expressions into $x^2+y^2=3z^2$ gives \begin{align*} (3a)^2+(3b)^2=3z^2 \end{align*} so \begin{align*} 9a^2+9b^2=3z^2 \end{align*} and hence \begin{align*} 3(a^2+b^2)=z^2. \end{align*} Thus $3\mid z^2$. Using again that the only square residues modulo $3$ are $0$ and $1$, the divisibility $3\mid z^2$ implies $z^2\equiv 0 \pmod{3}$, hence $z\equiv 0 \pmod{3}$. Therefore $3\mid z$. [/step] [step:Contradict the primitive condition] We have proved $3\mid x$, $3\mid y$, and $3\mid z$. Therefore $3$ is a common positive divisor of $x$, $y$, and $z$. This implies $\gcd(x,y,z)\ge 3$. This contradicts the assumed primitive condition $\gcd(x,y,z)=1$. Hence no integers $x,y,z\in \mathbb{Z}$ with $\gcd(x,y,z)=1$ satisfy $x^2+y^2=3z^2$. [/step]