Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5.
The condition is symmetric so we can assume that \( b \leq a \).

The first case is when \( a = b \). In this case, \( a! + a = 5^m \) for some positive integer \( m \). We can rewrite this as \( a \cdot ((a-1)! + 1) = 5^m \). This means that \( a = 5^k \) for some integer \( k \geq 0 \). It is clear that \( k \) cannot be 0. If \( k \geq 2 \), then \( (a-1)! + 1 = 5^l \) for some \( l \geq 1 \), but \( a-1 = 5^k - 1 > 5 \), so \( 5 \mid (a-1)! \), which is not possible because \( 5 \mid (a-1)! + 1 \). This means that \( k = 1 \) and \( a = 5 \). In this case, \( 5! + 5 = 125 \), which gives us the solution \( (5,5) \).

Let us now assume that \( 1 \leq b < a \). We have \( a! + b = 5^x \) and \( b! + a = 5^y \). Since \( b < a \), \( b! + a < a! + a \), so \( 5^y < 5^x \), which implies \( y < x \). If \( x \geq 2 \), then \( 5 \mid a! \). However, \( 5 \mid 5^y = a! + 1 \), which leads to a contradiction. We conclude that \( x = 1 \) and \( a = 4 \). From here \( a! + b = 25 \) and \( b! + a = 5 \), so we get two more solutions: \( (1,4) \) and \( (4,1) \).

Now we focus on the case \( 1 < b < a \). Because \( b \mid 5^x \) and \( b > 1 \), we have \( b = 5^z \) for \( z \geq 1 \). If \( z \geq 2 \), then \( 5b = 5 \), which gives us \( a \geq 10 \). However, this would mean that \( 25 \mid a! \), \( 5 \mid b \) and \( 25 \nmid b \), which is not possible, because \( a! + b = 5^x \) and \( 25 \mid 5^x \).

We conclude that the only solutions are \( (1,4) \), \( (4,1) \) and \( (5,5) \).