, 2013

Suppose such two numbers $a$ and $b$ exist.
Because $48 \mid 48(a+1)^b$, it also holds that $48 \mid a! + 48$ or $48 \mid a!$. Because $48 = 2^4 \cdot 3$, we have $48 \mid a!$ only for $a \ge 6$. The cases $a=6$ and $a=7$ are considered separately.

When $a=6$, we search for a solution of the equation $6! + 48 = 48 \cdot 7^b$. Hence $15+1=7^b$, which has no solutions in natural numbers since the left side is even and the right side is odd.

When $a=7$, we get the equation $7! + 48 = 48 \cdot 8^b$ or $105+1 = 8^b$. This equation also has no solutions because $106$ is not a power of $8$.

Hence it must hold true that $a \ge 8$. We now distinguish 2 cases:

* $a+1$ is not a prime number. Because $a \ge 8$, the equation can be divided by $48$ and written as
$$3 \cdot 5 \cdot 7 \cdot 8 \dots a+1 = (a+1)^b. \quad (1)$$
The number $3 \cdot 5 \cdot 7 \cdot 8 \dots a$ has the property that it is divisible by all prime numbers smaller than or equal to $a$. Now suppose $p$ is a prime number that divides $a+1$. Because $p+1$ is a composite number, we have $p \le a$. Hence $3 \cdot 5 \cdot 7 \cdot 8 \dots a \equiv 0 \pmod p$. Because we also have $(a+1)^b \equiv 0 \pmod p$, from (1) we derive $1 \equiv 0 \pmod p$. This is a contradiction.

* $a+1$ is a prime number. According to Wilson's theorem we now have $a! \equiv -1 \pmod{(a+1)}$. It also holds that $48(a+1)^b \equiv 0 \pmod{(a+1)}$, hence
$$0 \equiv a! + 48 \equiv -1 + 48 \equiv 47 \pmod{(a+1)}.$$
From this we get $a+1 \mid 47$ and, further, because $47$ is prime, $a+1=47$. We obtain the equation $46!+48 = 48 \cdot 47^b$ or $15 \cdot (7 \cdot 8 \dots 46) + 1 = 47^b$. The next estimate defines a range of possible values of $b$:
$$15 \cdot (7 \cdot 8 \dots 46) + 1 = 24 \cdot 35 \cdot \underbrace{(9 \cdot 10 \dots 46)}_{38 \text{ stevil}} + 1 < 47^{40}$$
Hence $b < 40$.

Since it holds modulo $41$ that $15 \cdot (7 \cdot 8 \dots 41 \dots 46) + 1 \equiv 1 \pmod{41}$, it must also hold that $47^b \equiv 6^b \equiv 1 \pmod{41}$. We now only need to check if such a number exists.

This can be done using Fermat's little theorem. The theorem gives $6^{40} \equiv 1 \pmod{41}$. Let now $c$ be the smallest exponent such that $6^c \equiv 1 \pmod{41}$. It is known that $c \mid 40$, so we just need to observe the divisors of $40$. Because $6^8 \equiv 10 \pmod{41}$ and $6^{20} \equiv -1 \pmod{41}$, neither of the divisors smaller than $40$ is good. The smallest solution is thus $c=40$, which means that equation $6^b \equiv 1 \pmod{41}$ has no solutions for $b < 40$.

Hence, the initial equation has no solutions in natural numbers.