\section*{Problem 5 - 181045}

Determine all pairs of natural numbers \((n ; z)\) for which \(2^{n}+12^{2}=z^{2}-3^{2}\) holds!
}

The given equation is equivalent to

\[
z^{2}-2^{n}=153
\]

(I) Suppose, for a pair of natural numbers \((N ; z)\), (1) is satisfied.

1. Case: \(n\) is even, i.e., \(n=2 m\) with natural \(m\). From (1) it follows then

\[
\left(z-2^{m}\right)\left(z+2^{m}\right)=153
\]

Since \(153=3^{2} \cdot 17\) has only the decompositions into two integer factors, where the first is smaller than the second and this (and thus also the first) is greater than 0, namely \(1 \cdot 153, 3 \cdot 51\) and \(9 \cdot 17\), there are at most the possibilities for (2)

\[
\begin{array}{cc}
z-2^{m}=1 & , \quad z+2^{m}=153 \\
z-2^{m}=3 & , \quad z+2^{m}=51 \\
z-2^{m}=9 & , \quad z+2^{m}=17
\end{array}
\]

Of these, (3) leads to the contradiction \(2^{m}=76\) and (4) to the contradiction \(2^{m}=24\); (5) leads to \(z=13, 2^{m}=4\), thus \(m=2, n=4\).

2. Case: \(n\) is odd. It holds \(2 \equiv 1(\bmod 3)\), thus \(2^{n} \equiv-1(\bmod 3)\). If now \(z \equiv 0(\bmod 3)\), it follows

\[
z^{2}-2^{n} \equiv 0+1(\quad \bmod 3)
\]

if \(z \equiv-1(\bmod 3)\), it follows

\[
z^{2}-2^{n} \equiv 1+1(\quad \bmod 3)
\]

Since \(153 \equiv 0(\bmod 3)\), both (6) and (7) result in a contradiction to (1). Therefore, (1) can only be satisfied by \((4 ; 13)\).

(II) Indeed, these numbers satisfy (1), because it holds \(2^{4}+12^{2}=160=169-9\). Thus, exactly this pair of numbers is the one sought.