Exercise 7. An integer $k>1$ is said to be superb if there exist $m, n$, and $a$, three strictly positive integers such that

$$
5^{\mathrm{m}}+63 \mathrm{n}+49=\mathrm{a}^{\mathrm{k}}
$$

Determine the smallest superb integer.
## Solution to Exercise 7

Suppose $k=2$ is superb: there exist three strictly positive integers $m, n$, and $a$ such that $5^{m} + 63n + 49 = a^{2}$. By looking modulo 3, we have $5^{m} + 1 \equiv 2^{m} + 1 \equiv a^{2} \pmod{3}$. The powers of 2 modulo 3 alternate between 2 if $m$ is odd and 1 if $m$ is even, so $2^{m} + 1$ is either 2 or 0 modulo 3. Since the squares modulo 3 are 0 and 1, we deduce that $2^{m} + 1$ must be 0 modulo 3, hence $m$ is odd. By looking modulo 7, we get $5^{m} \equiv a^{2} \pmod{7}$. The squares modulo 7 are $0, 1, 2, 4$, and the powers of 5 modulo 7 alternate between $1, 5, 4, 6, 2, 3$. Thus, $5^{m}$ is 0, 1, 2, or 4 if and only if $m$ is 0, 2, or 4 modulo 6, which is impossible since $m$ is odd. Therefore, $k \neq 2$.

Suppose $k=3$ is superb: there exist three strictly positive integers $m$, $n$, and $a$ such that $5^{m} + 63n + 49 = a^{3}$. By looking modulo 7, we have $5^{m} \equiv a^{3} \pmod{7}$. The cubes modulo 7 are $0, 1, 6$. The powers of 5 modulo 7 alternate between $1, 5, 4, 6, 2, 3$. Thus, we get that $m \equiv 0 \pmod{6}$ or $m \equiv 3 \pmod{6}$; in all cases, 3 divides $m$. We then look modulo 9: we have $5^{m} + 4 \equiv a^{3} \pmod{9}$. Since $m$ is divisible by 3, $5^{m}$ and $a^{3}$ are cubes. The cubes modulo 9 are $0, 1, 8$. Thus, $5^{m} - a^{3}$ can be 0, 1, 2, 7, 8 modulo 9, but not 4, which is contradictory. Therefore, $k \neq 3$.

Suppose $k=3$ is superb: there exist three strictly positive integers $m, n$, and $a$ such that $5^{m} + 63n + 49 = a^{4} = (a^{2})^{2}$. In particular, 2 is superb, which is contradictory.

Note that 5 is superb: by taking $a=3$, we have $a^{k} = 3^{5} = 243 = 5^{1} + 63 \times 3 + 49$, so 5 is superb. Thus, 5 is the smallest superb integer.

Comment from the graders: The exercise was attempted by few students, who solved it well.