## Task 4.

Determine all pairs $(a, b)$ of natural numbers for which

$$
a^{5}+a^{4}=7^{b}-1
$$
## First Solution.

Let's write the given equation as

$$
\left(a^{2}+a+1\right)\left(a^{3}-a+1\right)=7^{b}
$$

If $a=1$, then $7^{b}-1=2$, which is impossible. Therefore, $a>1$, which means $a^{3}-a+1>1$, so there exist natural numbers $k$ and $n$ such that

$$
a^{2}+a+1=7^{k} \quad \text { and } \quad a^{3}-a+1=7^{n}
$$

We conclude that $7 \mid a^{3}-a+1$. By checking all possible remainders when divided by 7, we conclude that $7 \mid a-2$. Therefore, there exists a non-negative integer $c$ such that $a=7 c+2$. Substituting this, we get

$$
7 c^{2}+5 c+1=7^{k-1} \quad \text { and } \quad 49 c^{3}+42 c^{2}+11 c+1=7^{n-1}
$$

If $k=1$, then $c=0$, so $a=2$, which gives the solution $(a, b)=(2,2)$.

If $k>1$, then it is necessary that $n>1$. From the first equation, we have $7 \mid 5 c+1$, and from the second, $7 \mid 4 c+1$, which is impossible.

Therefore, the only solution is $(a, b)=(2,2)$.