Turkey — Team Selection Test

Note that $n \ge 5$ and $m$ and $p$ have the same parity. If $m$ is even then $p = 2$ and we get $0 \equiv 4 \pmod{8}$. Therefore, $m$ and $p$ both are odd. $m^3 \equiv 0, \pm1 \pmod{7}$ and $2^n \equiv 2, 4, 1 \pmod{7}$. Therefore, $2^n \equiv 1 \pmod{7}$ and consequently $n = 3k$: $2^{3k} - m^3 = 7p^2$. We get $(2^k - m)(2^{2k} + 2^k m + m^2) = 7p^2$.

Let $\gcd(2^k - m, 2^{2k} + 2^k m + m^2) = d$. Since $m$ is odd $d$ also is odd. Since $d$ divides $(2^k - m)^2$ and $2^{2k} + 2^k m + m^2$, $d$ divides $3 \cdot 2^k m$. If $d$ divides $m$ then since $d$ also divides $2^k - m$ we get that $d$ divides $2^k$ and consequently $d = 1$. If $d$ does not divide $m$ then $d = 3$.

Note that $(2^k - m < 2^{2k} + 2^k m + m^2)$.

Case 1: $d = 1$ and $2^k - m = 1$, $2^{2k} + 2^k m + m^2 = 7p^2$. Plugging $m = 2^k - 1$ into $2^{2k} + 2^k m + m^2 = 7p^2$ we get $3 \cdot 2^{2k} - 3 \cdot 2^k + 1 - 7p^2 = 0$. Therefore $k > 1$ and $1 + p^2 \equiv 0 \pmod{4}$. No solution in this case.

Case 2: $d = 1$ and $2^k - m = 7$, $2^{2k} + 2^k m + m^2 = p^2$. Plugging $m = 2^k - 7$ into $2^{2k} + 2^k m + m^2 = p^2$ we get $3 \cdot 2^{2k} - 21 \cdot 2^k + 49 = p^2$. Since $d = 1$ $p \ne 7$. In $\pmod{7}$ left hand side takes $3 \cdot \{2, 4, 1\} = \{6, 5, 3\}$ and right hand side takes $\{1, 2, 4\}$. No solution in this case.

Case 3: $d = 3$ and $2^k - m = p$, $2^{2k} + 2^k m + m^2 = 7p$. Readily $p = d = 3$. Plugging $2^k = m + 3$ into $2^{2k} + 2^k m + m^2 = 21$ we get $3m^2 + 9m - 12 = 0$. Therefore $m = 1$ or $m = -4$. Since $m$ is nonnegative we get $m = 1$. Therefore, $k = 2$ and $n = 6$. The only solution is $(6, 1, 3)$.