4. Determine the rational values of $x$ for which the number $8 x^{2}-2 x-3$ is the square of a rational number.
Solution. Let's factor the given expression:

$$
8 x^{2}-2 x-3=(2 x+1)(4 x-3)
$$

For rational values of $x$, the two factors on the right side are rational numbers. The product of these two rational numbers is a square of a rational number if and only if there exist rational numbers $d, u$, and $v$ such that:

$$
2 x+1=d u^{2} \text { and } 4 x-3=d v^{2}
$$

Solving (2) as a system of equations for unknowns $x$ and $d$ yields

$$
d=\frac{5}{2 u^{2}-v^{2}} \text { and } x=\frac{1}{2} \frac{3 u^{2}+v^{2}}{2 u^{2}-v^{2}} .
$$

Since $u$ and $v$ are rational, and $\sqrt{2}$ is irrational, the expression that appears as the denominator is 0 if and only if $u=v=0$. Therefore, the only invalid choice for $u$ and $v$ is $u=v=0$. For any other choice of rational numbers $u$ and $v$, the rational number $x$ (chosen as in (3)) satisfies the requirements of the problem, and it holds that

$$
8 x^{2}-2 x-3=\left(\frac{5 u v}{2 u^{2}-v^{2}}\right)^{2}
$$

## III year