Exercise 12. Determine all triplets of positive integers $(\mathrm{a}, \mathrm{b}, \mathrm{n})$ satisfying:

$$
a!+b!=2^{n}
$$
Solution to Exercise 12: Since factorials have many common odd factors, we immediately realize that obtaining a power of 2 will be very restrictive.

Suppose that $\mathrm{a}, \mathrm{b} \geqslant 3$, then 3 divides the sum of the factorials and thus $2^{\mathrm{n}}: c^{\prime}$ is absurd. This means that one of the two must be in $\{1,2\}$. By symmetry, we only need to consider two cases:

1. $a=1$: there are two small cases:

- $\mathrm{b}=1: \mathrm{n}=1$

- $\mathrm{b} \geqslant 2$: the two sides do not have the same parity, which is impossible

2. $a=2: b \geqslant 2$ (as seen above), a few cases:

- $\mathrm{b}=2,3: \mathrm{n}=2,3$

- $\mathrm{b} \geqslant 4$: there is a problem modulo 4, which is impossible

Therefore: the only solutions are in $\{(1,1,1),(2,2,2),(2,3,3),(3,2,3)\}$.

Grader's Comment

There were quite a few logical errors. It is not because one cannot have $a \geqslant 3$ and $b \geqslant 3$ at the same time that one must have $\mathrm{a}<3$ and $\mathrm{b}<3$ (rather, $\mathrm{a}<3$ or $\mathrm{b}<3$). 1 is a power of 2 that should not be forgotten, and 0 is not in $\mathbb{N}^{*}$, and 1 is an odd divisor of 2 that should not be forgotten either. Do not forget to mention the symmetric solutions if you assume $\mathrm{a} \geqslant \mathrm{b}$.