Problem 1539

Solution:
We cannot have $c=0$ as $a!+5^{b} \geqslant 2 > 1 = 7^{0}$.

Assume first that $b=0$. So we are solving $a!+1=7^{c}$. If $a \geqslant 7$, then $7 \mid a!$ and so $7 \nmid a!+1$. So $7 \nmid 7^{c}$ which is impossible as $c \neq 0$. Checking $a<7$ by hand, we find the solution $(a, b, c) = (3, 0, 1)$.

We now assume that $b>0$. In this case, if $a \geqslant 5$, we have $5 \mid a!$, and since $5 \mid 5^{b}$, we have $5 \mid 7^{c}$, which obviously cannot be true. So we have $a \leqslant 4$. Now we consider the following cases:

Case 1. Suppose $a=0$ or $a=1$. In this case, we are solving the equation $5^{b}+1=7^{c}$. However the Left Hand Side of the equation is always even, and the Right Hand Side is always odd, implying that this case has no solutions.

Case 2. Suppose $a=2$. Now we are solving the equation $5^{b}+2=7^{c}$. If $b=1$, we have the solution $(a, b, c) = (2, 1, 1)$. Now assume $b \geqslant 2$. We have $5^{b}+2 \equiv 2 \pmod{25}$ which implies that $7^{c} \equiv 2 \pmod{25}$. However, by observing that $7^{4} \equiv 1 \pmod{25}$, we see that the only residues that $7^{c}$ can have when divided with 25 are $7, 24, 18, 1$. So this case has no more solutions.

Case 3. Suppose $a=3$. Now we are solving the equation $5^{b}+6=7^{c}$. We have $5^{b}+6 \equiv 1 \pmod{5}$ which implies that $7^{c} \equiv 1 \pmod{5}$. As the residues of $7^{c}$ modulo 5 are $2, 4, 3, 1$, in that order, we obtain $4 \mid c$.

Viewing the equation modulo 4, we have $7^{c} \equiv 5^{b}+6 \equiv 1+2 \equiv 3 \pmod{4}$. But as $4 \mid c$, we know that $7^{c}$ is a square, and the only residues that a square can have when divided by 4 are 0, 1. This means that this case has no solutions either.

Case 4. Suppose $a=4$. Now we are solving the equation $5^{b}+24=7^{c}$. We have $5^{b} \equiv 7^{c}-24 \equiv 1-24 \equiv 1 \pmod{3}$. Since $5 \equiv 2 \pmod{3}$, we obtain $2 \mid b$. We also have $7^{c} \equiv 5^{b}+24 \equiv 4 \pmod{5}$, and so we obtain $c \equiv 2 \pmod{4}$. Let $b=2m$ and $c=2n$. Observe that
$$24 = 7^{c} - 5^{b} = (7^{n} - 5^{m})(7^{n} + 5^{m})$$
Since $7^{n} + 5^{m} > 0$, we have $7^{n} - 5^{m} > 0$. There are only a few ways to express $24 = 24 \cdot 1 = 12 \cdot 2 = 8 \cdot 3 = 6 \cdot 4$ as a product of two positive integers. By checking these cases we find one by one, the only solution in this case is $(a, b, c) = (4, 2, 2)$.

Having exhausted all cases, we find that the required set of triples is
$$(a, b, c) \in \{(3, 0, 1), (2, 1, 1), (4, 2, 2)\}$$