\section*{Problem 4 - 201034}

Determine all triples \((a, h, x)\) of non-zero natural numbers with the following property! If \(a\) and \(h\) are the measurements in centimeters of the base edge length and height, respectively, of a right square pyramid, then both the surface area in square centimeters and the volume in cubic centimeters of this pyramid have the measurement \(x\).
}

The volume is

\[
x=\frac{1}{3} a^{2} h
\]

The surface area is

\[
x=a^{2}+4 \cdot\left(\frac{1}{2} a \sqrt{h^{2}+\frac{1}{4} a^{2}}\right)=a^{2}+a \sqrt{a^{2}+4 h^{2}}
\]

Thus, it must hold that:

\[
\frac{1}{3} a^{2} h=a^{2}+a \sqrt{a^{2}+4 h^{2}}
\]

Both sides of the equation must be integers. We can divide both sides by \(a\), and both sides must still be integers:

\[
\frac{1}{3} a h=a+\sqrt{a^{2}+4 h^{2}}
\]

(Justification: \(a^{2} h\) must be divisible by 3. Either \(a\) is a multiple of 3, in which case \(\frac{1}{3} a h\) remains an integer, or \(h\) is divisible by 3, in which case \(\frac{1}{3} a h\) is definitely an integer). We first solve the equation for \(a\):

\[
\begin{aligned}
& \left(\frac{1}{3} h-1\right) a=\sqrt{a^{2}+4 h^{2}} \quad \rightarrow \quad \frac{1}{9} h^{2} a^{2}-\frac{2}{3} h a^{2}+a^{2}=a^{2}+4 h^{2} \\
& a^{2}=\frac{4 h}{\frac{1}{9} h-\frac{2}{3}}=\frac{36 h}{h-6} \quad \rightarrow \quad a=6 \sqrt{\frac{h}{h-6}}
\end{aligned}
\]

Thus, \(\frac{h}{h-6}\) must be a perfect square. Clearly, \(h > 6\) must hold, but also

\[
\frac{h}{h-6} \geq 4
\]

since \(\frac{h}{h-6}=1\) is not possible. This implies: \(h \geq 4 h-24\) and \(h \leq 8\).

Therefore, the only possible values for \(h\) are 7 and 8, but for \(h=7\) no perfect square exists. Thus, \(h=8\), and by substitution, we get \(a=12\) and \(x=384\). The only triplet that satisfies the conditions is therefore \((a ; h ; x)=(12 ; 8 ; 384)\).