Let's determine those real numbers $p$ and $q$ for which $2^{p x+q}=p \cdot 2^{x}+q$ for all real $x$.

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Determine the real numbers $p$ and $q$ such that $2^{p x+q}=p \cdot 2^{x}+q$ for all real $x$.
Since the equality holds for all real $x$, it also holds for $x=0 ; x=1 ; x=2$. Substituting these into the given relationship, we get the following three equations:

$$
\begin{aligned}
2^{q} & =p+q \\
2^{p+q} & =2 p+q \\
2^{2 p+q} & =4 p+q
\end{aligned}
$$

Subtract the first equation from the second, and the second from the third:

$$
\begin{aligned}
2^{q}\left(2^{p}-1\right) & =p \\
2^{p+q}\left(2^{p}-1\right) & =2 p
\end{aligned}
$$

First, assume that $2^{p} \neq 1$, i.e., $p \neq 0$. Then, dividing equation (5) by equation (4) gives $2^{p}=2$, from which we get $p=1$. Substituting $p=1$ into equations (1) and (2) yields:

$$
2^{q}=1, \quad \text { so } \quad q=0
$$

The value pair $p=1, q=0$ is indeed a solution, as in this case the original relationship is an identity.

We need to separately examine the case $p=0$. In this case, both sides of the examined equality are independent of $x$, and our task is to solve the equation:

$$
2^{q}=q
$$

We claim that for all real $q$, $2^{q}>q$, so there is no solution other than $q=0$. For this, we need the following inequality:

$$
2^{n} \geq n+1, \quad \text { if } n \text { is an integer. }
$$

The above inequality clearly holds if $n \geq [q]$, where $[q]$ is the greatest integer not greater than $q$. Thus, $2^{q} \geq 2^{[q]} \geq [q]+1 > q$, so indeed there is no further solution when $p=0$.

Thus, the solution to the problem is: $p=1 ; q=0$.

Alberti Gábor (Budapest, Árpád Gym., II. o. t.)

Remark. We present another way to clarify the case $p \neq 0$. It is known that the range of the function $f(x)=2^{x}$ is the set of positive real numbers. If $p \neq 0$, then $px+q$ takes on all real values, so the range of the function $g(x)=2^{px+q}$ is also the set of positive real numbers. The range of the function on the right side is the set of real numbers greater than or less than $q$, depending on whether $p$ is positive or negative. Since the two functions are equal, their ranges are also the same, which means that $p$ is positive and $q=0$, from which $2^{px}=p \cdot 2^{x}$ follows. Since the equality holds for $x=0$, we have $2^{0}=p$, so $p=1$.