Find all integer solutions $x$ of the equation: $x^{3}+(x+1)^{3}+(x+2)^{3}=(x+3)^{3}$
By developing the left-hand side, we have: $3 x^{3}+9 x^{2}+15 x+9=(x+3)^{3}$. It follows that $x+3$ is divisible by 3, and thus $x$ is also divisible by 3, and we can set: $x=3 t$, which leads us to: $81 t^{3}+81 t^{2}+45 x+9=27\left(t^{3}+3 t^{2}+3 t+1\right)$, or, after simplifications:
$3 t^{3}-2 t-1=0$. 1 is an obvious root, and we can factorize: $3 t^{3}-2 t-1=(t-1)\left(3 t^{2}+3 t+1\right)$. The term $3 t^{2}+3 t+1$ being always strictly positive, this equation admits for its unique root (integer or not): $t=1$, so the initial equation admits for its unique root, a fortiori for its unique integer root: $x=3$. The relation; $3^{3}+4^{3}+5^{3}=6^{3}$ is classic.

## - Reminder on congruences -

Two integers $a$ and $b$ are congruent modulo $n: a \equiv b(\bmod n)$ if their difference is divisible by $n$. This is an equivalence relation, allowing all integers to be divided into $n$ congruence classes: those of the form $n k, n k+1, \ldots n k+(n-1)$. These classes can be manipulated almost like numbers, with additional properties, notably Fermat's (little) theorem: if $p$ is a prime number and $a$ is an integer not divisible by $p, a^{p-1} \equiv 1(\bmod p)$. This can be proven by noting that modulo $p, a, 2 a, \ldots(p-1) a$ are each in a different class, so $a \times 2 a \times \ldots \times(p-1) a \equiv 1 \times 2 \times \ldots \times(p-1)$ (since on both sides, we find all non-zero classes modulo $p$), which simplifies.