\section*{Exercise 4 - 241044}

Determine all pairs \((a ; b)\) of natural numbers for which \(a! + b! = (a + b)!\) holds! Hint: For every natural number \(n \geq 1\), \(n!\) is defined as the product of all natural numbers \(k\) such that \(1 \leq k \leq n\); furthermore, \(0! = 1\) is defined.
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Due to symmetry, it generally holds that \((b ; a)\) is a solution if \((a ; b)\) is a solution. We therefore consider only cases where \(a \leq b\). Since \(0!=1\), it is clear that \(a>0\) must hold, because otherwise \(1+b!=b\) ! would have to hold, which is not possible. The factorial function can be represented as

\[
n!=\prod_{k=1}^{n} k
\]

Generally, it should hold that: \(a!+b!=(a+b)\) !

\[
\prod_{k=1}^{a} k+\prod_{k=1}^{b} k=\prod_{k=1}^{a+b} k
\]

We divide by \(b!:\)

\[
\frac{\prod_{k=1}^{a} k}{\prod_{k=1}^{b} k}+1=\frac{\prod_{k=1}^{a+b} k}{\prod_{k=1}^{b} k}
\]

Since \(b \geq a\) and \((a+b)>b\), we can simplify:

\[
\frac{1}{\prod_{k=a+1}^{b} k}+1=\prod_{k=b+1}^{a+b} k
\]

On the right, we have a natural number, so the fraction on the left must also result in a natural number. This is not possible for \(b>a\). Therefore, only the cases \(a=b>0\) remain to be investigated. Then it must hold that

\[
2 \prod_{k=1}^{a} k=\prod_{k=1}^{2 a} k \quad \rightarrow \quad 2=\frac{\prod_{k=1}^{2 a} k}{\prod_{k=1}^{a} k} \quad \rightarrow \quad 2=\prod_{k=a+1}^{2 a} k
\]

This is only satisfied for \(a=1\), for larger \(a\) it always holds that

\[
2 \cdot a!<(2 a)!
\]

Therefore, the pair \((1 ; 1)\) is the only solution.