Exercise 6. Find all pairs of strictly positive integers $(\mathrm{m}, \mathrm{n})$ for which:

$$
1+2^{\mathfrak{n}}+3^{n}+4^{n}=10^{m}
$$
Solution to Exercise 6 This equation is valid for all $\mathrm{n}, \mathrm{m}$, so it is also valid when taken modulo an integer $\mathrm{k}$, that is, by only considering the remainders of the division by $\mathrm{k}$. We start by looking at it modulo 3:

$$
1+(-1)^{\mathrm{n}}+0+1^{\mathrm{n}} \equiv 1^{\mathrm{m}}(\bmod 3) \text{ so } (-1)^{\mathrm{n}} \equiv -1
$$

Therefore, $\mathrm{n}$ is odd: let $\mathrm{k} \in \mathbb{N}$ such that $\mathrm{n}=2 \mathrm{k}+1$.

We now assume that $\mathrm{n}, \mathrm{m} \geqslant 3$ and look at it modulo 8:

$$
1+0+3 \cdot 3^{2 k}+0 \equiv 0(\bmod 8) \text{ i.e. } 1+3 \cdot 1 \equiv 0
$$

This is absurd.

We deduce that one of the two is in $\{1,2\}$.

It then suffices to consider the cases $n=1, n=2, m=1$ and $m=2$.

For $\boldsymbol{n}=1$ we find $1+2+3+4=10$, so $(1,1)$ is a solution. For $\boldsymbol{n}=2, 1+4+9+16=30$ is not a power of 10 (since 30 is divisible by 3). Since the function that maps $n$ to $1+2^{n}+3^{n}+4^{n}$ is strictly increasing for a fixed $\mathrm{m}$, there is at most one solution. Since $(1,1)$ and $(3,2)$ are solutions (because $1+2^{3}+3^{3}+4^{3}=1+8+27+64=100=10^{2}$), these are the only solutions with $\mathrm{m}=1$ or 2. The set of solutions is therefore $\{(1,1),(2,3)\}$.

Grader Comments

Even though many students almost succeeded in the exercise, few achieved the maximum score: indeed, several did not justify why, for $\mathrm{m}=2$, only $\mathrm{n}=3$ was a solution (which could have been done by bounding or using a growth argument). Writing is not a race; it is better to write a page and have all the arguments well explained than to compact everything into 9 lines and end up with a poorly clear and often incomplete paper (moreover, compact papers are often much harder to read than longer but well-detailed papers). Do not forget to verify that the pairs obtained do indeed satisfy the equation.