Exercise 4. Determine all triplets of positive integers $(\mathrm{a}, \mathrm{b}, \mathrm{n})$ satisfying:

$$
a!+b!=2^{n}
$$
Solution to Exercise 4 Since factorials have many common odd factors, we immediately realize that obtaining a power of 2 will be very restrictive.

Suppose that $\mathrm{a}, \mathrm{b} \geqslant 3,3$ divides the sum of the factorials and thus $2^{\mathrm{n}}: \mathrm{c} \text{ is absurd}$. That is, 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 $\{(1,1,1),(2,2,2),(2,3,3),(3,2,3)\}$.

Graders' Comments

The exercise was generally well done. It is important not to forget to specify that we can assume $\mathrm{a} \geqslant \mathrm{b}$ before using this inequality in the reasoning.