Let's determine the positive integer parameters $t, u, v$ such that the function

$$
y=f(x)\left(2^{3 u-5 v-3}-1\right) x^{2}+\left(2^{7 u-8 v-18}-2^{4 u-3 v-15}\right) x+\frac{11 t^{2}-5 u}{154 v-10 t^{2}}
$$

has a positive minimum at $x=-\frac{1}{8}$.
We know that the quadratic function $y=a x^{2}+b x+c$ has a minimum at the point $x=-\frac{b}{2 a}$, provided that $a>0$. In this case, the location of the extremum is

$$
x=-\frac{b}{2 a}=-\frac{1}{8}
$$

from which

$$
a=4 b
$$

or

$$
2^{3 u-5 v-3}-1=4\left(2^{7 u-8 v-18}-2^{4 u-3 v-15}\right)
$$

From this,

$$
2^{3 u-5 v-3}-1=2^{7 u-8 v-16}-2^{4 u-3 v-13}=2^{4 u-3 v-13}\left(2^{3 u-5 v-3}-1\right)
$$

or, reducing the equation to 0,

$$
\left(2^{3 u-5 v-3}-1\right)\left(2^{4 u-3 v-13}-1\right)=0
$$

It follows that one of the factors on the left side must be zero. However, the first factor cannot be zero, because in the function $y=f(x)$ it is the coefficient of the quadratic term.

Therefore, it must be that

$$
2^{4 u-3 v-13}-1=0
$$

and from this (since the only zeroth power of 2 is 1),

$$
4 u-3 v-13=0
$$

The solution to this Diophantine equation is

$$
v=u-4+\frac{u-1}{3}=u-4+z
$$

where

$$
z=\frac{u-1}{3}
$$

and thus

$$
\begin{aligned}
& u=3 z+1, \text { and } \\
& v=4 z-3
\end{aligned}
$$

(Since $u$ and $v$ are positive integer parameters

$$
\begin{aligned}
u & =3 z+1>0 \\
v & =4 z-3>0 \\
(3) \text {-from } z>-\frac{1}{3}, & \text { (4)-from } z>\frac{3}{4}
\end{aligned}
$$

the latter includes the former.

On the other hand, since the problem specifies a minimum, the coefficient of the quadratic term must be positive, i.e.,

$$
2^{3 u-5 v-3}-1>0
$$

which holds only if

$$
3 u-5 v-3>0
$$

Substituting the values of $u$ and $v$ from (1) and (2), we get

$$
9 z+3-20 z+15-3>0
$$

or

$$
z<\frac{15}{11}
$$

and from (4),

$$
z>\frac{3}{4}
$$

Thus, the only integer value for $z$ is 1.

Substituting $z=1$ into (1) and (2), we get

$$
u=4, \quad v=1
$$

Finally, we need to check the condition for the minimum, which is

$$
2^{3 u-5 v-3}-1>0
$$

Substituting $u=4$ and $v=1$, we get

$$
2^{3 \cdot 4 - 5 \cdot 1 - 3} - 1 = 2^4 - 1 = 15 > 0
$$

Thus, the function is

$$
y=f(x)=15 x^{2}+\frac{15}{4} x+\frac{79}{64}
$$

Károly Bartók (Székesfehérvár, József A. g. II. o. t.)