Solve the following equation:

$$
\sqrt{4^{x}+\frac{17}{64}}-\sqrt{2 \cdot 4^{x}-\frac{7}{64}}=\sqrt{4^{x}-\frac{1}{16}}
$$
I. Solution: Square both sides of the equation:

$$
4^{x}+\frac{17}{64}-2 \sqrt{2 \cdot 4^{2 x}+\left(\frac{34}{64}-\frac{7}{64}\right) 4^{x}-\frac{7 \cdot 17}{64^{2}}}+2 \cdot 4^{x}-\frac{7}{64}=4^{x}-\frac{1}{16}
$$

After simplification and rearrangement:

$$
2 \cdot 4^{x}+\frac{14}{64}=2 \sqrt{2 \cdot 4^{2 x}+\frac{27}{64} \cdot 4^{x}-\frac{119}{64^{2}}}
$$

Dividing the equation by 2 and squaring again for $4^{x}$, we obtain the following quadratic equation:

$$
4^{2 x}+\frac{13}{64} \cdot 4^{x}-\frac{168}{64^{2}}=0
$$

From this, using the quadratic formula:

$$
4^{x}=\frac{-\frac{13}{64} \pm \sqrt{\left(\frac{13}{64}\right)^{2}+\frac{4 \cdot 168}{64^{2}}}}{2}=\frac{-13 \pm \sqrt{169+672}}{128}=\frac{-13 \pm 29}{128}
$$

The negative root is not a solution to the problem, as $4^{x}$ is never negative for any $x$. Therefore, the only solution is

$$
4^{x}=\frac{-13+29}{128}=\frac{1}{8}=\frac{1}{4^{\frac{3}{2}}}=4^{-\frac{3}{2}}
$$

From this, we get $x=-\frac{3}{2}$.

Substituting the obtained root back into the original equation, we can verify that it indeed satisfies the equation.

$$
\text{Attila Meskó (Bp. VII., Madách g. IV. o. t.) }
$$

II. Solution: Denote the square root terms in the equation as $u, v, t$:

$$
\sqrt{4^{x}+\frac{17}{64}}=u, \quad \sqrt{4^{x}-\frac{4}{64}}=v, \quad \sqrt{2 \cdot 4^{x}-\frac{7}{64}}=t
$$

It is easy to see that the following relationships hold among the auxiliary variables:

$$
\begin{gathered}
u^{2}-v^{2}=\frac{21}{64} \\
z^{2}+v^{2}=t^{2}+\frac{20}{64}
\end{gathered}
$$

Rearranging the original equation, we get:

$$
u-v=t
$$

Substituting the value of $t$ from (2) into (3):

$$
u^{2}+v^{2}=u^{2}-2 u v+v^{2}+\frac{20}{64}
$$

After rearrangement:

$$
u v=\frac{10}{64}
$$

Thus,

$$
u^{2} v^{2}=\frac{100}{64^{2}}
$$

Substituting the value of $u^{2}$ from (1) into the above equality, we obtain the following quadratic equation for $v^{2}$:

$$
\left(v^{2}\right)^{2}+\frac{21}{64} v^{2}-\left(\frac{10}{64}\right)^{2}=0
$$

From this,

$$
v^{2}=\frac{-\left(\frac{21}{64}\right) \pm \sqrt{\left(\frac{21}{64}\right)^{2}+4 \cdot\left(\frac{10}{64}^{2}\right)}}{2}=\frac{-21 \pm \sqrt{441+400}}{128}=\frac{-21 \pm 29}{128}
$$

$v^{2}$ cannot be negative, so $v^{2}=\frac{8}{128}$.

Using the original value of $v^{2}$:

$$
v^{2}=4^{x}-\frac{4}{64}=\frac{8}{128}
$$

Thus,

$$
4^{x}=\frac{8}{64}=\frac{1}{8}=4^{-\frac{3}{2}},
$$

and therefore,

$$
x=-\frac{3}{2}
$$

László Bognár (Veszprém, Lovassy g. III. o. t.)