[proofplan] We use the rational point $(-1,0)$ on the unit circle as a base point and intersect the circle with rational-slope lines through that point. First we verify directly that every displayed parametrized point lies on the circle. Then, given any rational point other than $(-1,0)$, we define the rational slope $t = Y/(X+1)$ of the line joining it to $(-1,0)$ and solve the resulting quadratic equation in $X$. The factor corresponding to the known intersection point $X=-1$ leaves the second intersection, which is exactly the displayed formula. [/proofplan] [step:Verify that the displayed rational functions give points on the unit circle] Let $t \in \mathbb{Q}$. Since $1+t^2 \neq 0$, define rational numbers $X_t,Y_t \in \mathbb{Q}$ by \begin{align*} X_t &= \frac{1-t^2}{1+t^2}, & Y_t &= \frac{2t}{1+t^2}. \end{align*} Then \begin{align*} X_t^2 + Y_t^2 &= \left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 \\ &= \frac{(1-t^2)^2 + 4t^2}{(1+t^2)^2} \\ &= \frac{1 - 2t^2 + t^4 + 4t^2}{(1+t^2)^2} \\ &= \frac{1 + 2t^2 + t^4}{(1+t^2)^2} \\ &= 1. \end{align*} Thus every point in the displayed parametrized family is a rational point on $X^2+Y^2=1$. Also $(-1,0) \in \mathbb{Q}^2$ and \begin{align*} (-1)^2 + 0^2 = 1, \end{align*} so the listed points all lie on the rational unit circle. [/step] [step:Attach a rational slope to every rational point other than the base point] Let $(X,Y) \in \mathbb{Q}^2$ satisfy $X^2+Y^2=1$, and assume $(X,Y) \neq (-1,0)$. Then $X \neq -1$: if $X=-1$, the equation gives \begin{align*} 1 + Y^2 = 1, \end{align*} so $Y^2=0$ and hence $Y=0$, contradicting $(X,Y) \neq (-1,0)$. Define \begin{align*} t := \frac{Y}{X+1}. \end{align*} Since $X,Y \in \mathbb{Q}$ and $X+1 \neq 0$, we have $t \in \mathbb{Q}$. By definition of $t$, \begin{align*} Y = t(X+1). \end{align*} [/step] [step:Solve the circle-line intersection and discard the known base-point root] Substituting $Y=t(X+1)$ into $X^2+Y^2=1$ gives \begin{align*} X^2 + t^2(X+1)^2 &= 1. \end{align*} Expanding and collecting powers of $X$, \begin{align*} (1+t^2)X^2 + 2t^2X + (t^2-1) &= 0. \end{align*} The left-hand side factors as \begin{align*} (1+t^2)X^2 + 2t^2X + (t^2-1) &= (X+1)\bigl((1+t^2)X + (t^2-1)\bigr). \end{align*} Since $X \neq -1$, the factor $X+1$ is nonzero. Therefore \begin{align*} (1+t^2)X + (t^2-1) &= 0. \end{align*} Because $1+t^2 \neq 0$, this gives \begin{align*} X = \frac{1-t^2}{1+t^2}. \end{align*} Using $Y=t(X+1)$, \begin{align*} Y &= t\left(\frac{1-t^2}{1+t^2}+1\right) \\ &= t\left(\frac{1-t^2+1+t^2}{1+t^2}\right) \\ &= \frac{2t}{1+t^2}. \end{align*} Thus every rational point on the unit circle different from $(-1,0)$ has the displayed parametrized form. [guided] We now recover the parameter from an arbitrary rational point and show that the algebra forces the displayed formula. The point $(-1,0)$ is special because we are drawing the line through it and the point $(X,Y)$. Since $(X,Y) \neq (-1,0)$, the previous step proved $X+1 \neq 0$, so the slope \begin{align*} t := \frac{Y}{X+1} \end{align*} is a well-defined rational number. This means precisely that $(X,Y)$ lies on the line through $(-1,0)$ with slope $t$: \begin{align*} Y = t(X+1). \end{align*} We substitute this line equation into the circle equation. Since $X^2+Y^2=1$ and $Y=t(X+1)$, we get \begin{align*} X^2 + t^2(X+1)^2 &= 1. \end{align*} Expanding the square gives \begin{align*} X^2 + t^2(X^2+2X+1) &= 1, \end{align*} and collecting terms on the left gives \begin{align*} (1+t^2)X^2 + 2t^2X + (t^2-1) &= 0. \end{align*} This quadratic has one root already known geometrically: the line was chosen to pass through $(-1,0)$, whose $X$-coordinate is $-1$. Algebraically, that root appears through the factorization \begin{align*} (1+t^2)X^2 + 2t^2X + (t^2-1) &= (X+1)\bigl((1+t^2)X + (t^2-1)\bigr). \end{align*} Our point is not the base point, so $X \neq -1$ and therefore $X+1 \neq 0$. Hence the second factor must vanish: \begin{align*} (1+t^2)X + (t^2-1) &= 0. \end{align*} Since $1+t^2 \neq 0$ for $t \in \mathbb{Q}$, we may divide by $1+t^2$ and obtain \begin{align*} X = \frac{1-t^2}{1+t^2}. \end{align*} Finally, substituting this value into the line equation gives \begin{align*} Y &= t(X+1) \\ &= t\left(\frac{1-t^2}{1+t^2}+1\right) \\ &= t\left(\frac{1-t^2+1+t^2}{1+t^2}\right) \\ &= \frac{2t}{1+t^2}. \end{align*} So the rational point $(X,Y)$ is exactly the parametrized point associated to the rational parameter $t$. [/guided] [/step] [step:Combine both inclusions to identify the full rational solution set] The first step proves that every listed point belongs to \begin{align*} \{(X,Y) \in \mathbb{Q}^2 : X^2+Y^2=1\}. \end{align*} The preceding steps prove the reverse inclusion: every rational solution is either the base point $(-1,0)$ or has the displayed form for the rational number $t=Y/(X+1)$. Therefore \begin{align*} \{(X,Y) \in \mathbb{Q}^2 : X^2 + Y^2 = 1\} = \{(-1,0)\} \cup \left\{ \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) : t \in \mathbb{Q} \right\}. \end{align*} This proves the rational parametrization of the unit circle. [/step]