Rearranging the equation we get pⁿ = m² − 3600 = (m − 60)(m + 60).

Clearly, both m - 60 and m + 60 must be factors of pⁿ. Since p is prime, the only factors of pⁿ are 1 and integer powers of p, so both brackets can only be 1 or an integer power of p.

If m − 60 = 1, then m + 60 = 121, giving the solution p = 11, n = 2 and m = 61. If m  60 ≠ 1, then p divides m - 60 and m + 60, so p divides the difference between them, 120.

Therefore p can be only 2, 3 or 5.

If p = 2, we are looking for two powers of 2 which differ by 120. Since the larger one must be at least 120 and beyond 128 the difference between powers of 2 is larger than 120, the only solution is m+ 60 = 128 and m − 60 = 8. This gives a solution of p = 2, n = 10 and m = 68.

If p = 3, we are looking for two powers of 3 which differ by 120. Since the larger one must be at least 120 and beyond 81 the difference between powers of 3 is larger than 120, there is no solution in this case.

If p = 5, we are looking for two powers of 5 which differ by 120. Since the larger one must be at least 120 and beyond 125 the difference between powers of 5 is larger than 120, the only solution is m + 60 =125 and m − 60 =5. This gives a solution of p = 5, n = 4 and m = 65.

The solutions are (m, n, p) = (61, 2, 11), (65, 4, 5) or (68, 10, 2).

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