Find all triplets of strictly positive integers $(a, b, c)$ such that $6^{a}=1+2^{b}+3^{c}$
We notice that 3 divides $2^{b}+1$, so $b$ is odd. If $b=1,1+3^{c-1}=2 \cdot 6^{a-1}$, so $a=1$. Indeed, if $a>1,3$ divides $1+3^{c-1}$, which is absurd. Therefore, if $b=1$, we have the solution $(1,1,1)$. Otherwise, $b \geqslant 3$, so $a \geqslant 2$ (since $6<1+2^{3}$). Modulo 4, we get $1+3^{c} \equiv 0[4]$, so $c$ is odd. Modulo 8, we then have $1+2^{b}+3^{c} \equiv 4[8]$, so $6^{a} \equiv 4[8]$, so $a=2$. Thus $2^{b}+3^{c}=35$, and an exhaustive study gives the solutions $(2,3,3)$ and $(2,5,1)$.

Sometimes when we cannot find a modulo or factorization that allows us to advance on an arithmetic problem, we can resort to inequalities. This method partly relies on the following proposition:

## Proposition 1.

If $a$ and $b$ are integers such that $a$ divides $b$, then $|a| \leqslant|b|$

This property, though obvious, is nonetheless essential in solving many difficult exercises. Indeed, it can often be useful to combine several inequalities of this type to obtain an enclosure on a variable or an absurdity. An interesting point to note in this regard is that the property can always be refined: for example, if we know that $a \neq \pm b$, then $|a| \leqslant 2|b|$.

A second observation that can be made and which seems obvious but is equally useful is the following: there is no other square between two consecutive squares. It can of course be adapted with cubes, etc. Without further ado, here are some application exercises. In this section, exercises 3, 4, and 5 have been treated.