273. For which rational $x$ will the expression $3 x^{2}-$ $-5 x+9$ represent the square of a rational number?
273. First solution. The problem requires finding all rational solutions $(x, y)$ (where $y \geqslant 0$) of the equation $3 x^{2}-5 x+9 = y^{2}$ with two unknowns $x$ and $y$ (cf. problem cycle 5). Clearly, one solution is $x=0, y=3$. Let $x=x_{1}, y=y_{1}+3$; then we get:

$$
3 x_{1}^{2}-y_{1}^{2}-5 x_{1}-6 y_{1}=0
$$

For any solution $\left(x_{1}, y_{1}\right)$, different from $(0,0), \quad \frac{y_{1}}{x_{1}}=\frac{m}{n}$, where $m$ and $n$ are coprime integers, $y_{1}=x_{1} \cdot(m / n)$; hence, we have:

$$
3 x_{1}^{2}-\frac{m^{2}}{n^{2}} x_{1}^{2}-5 x_{1}-6 \frac{m}{n} x_{1}=0
$$

or, since $x \neq 0, x_{1}=\frac{5 n^{2}+6 m n}{3 n^{2}-m^{2}}$. The formula $x=\frac{5 n^{2}+6 m n}{3 n^{2}-m^{2}}$ gives the complete answer to the problem (the solution $x=0$ is obtained when $n=0$).

Second solution. If $(x, y)$ is a rational point on the second-order curve (hyperbola) $y^{2}=3 x^{2}-5 x+9$, then the ratio $k=\frac{y-3}{x}$ is rational (possibly infinite). On the other hand, if $k$ is rational, then the line $y=k x+3$ intersects the curve at two rational points: $(0,3)$ and $\left(\frac{6 k+5}{3-k^{2}}, \frac{3 k^{2}+5 k+9}{3-k^{2}}\right)$. Therefore, all rational points on the curve correspond to such $x$ that $x=\frac{6 k+5}{3-k^{2}}$.