Найдите всех натуральных

такие, что

.
Пусть

и поэтому

и

, где

. Тогда делимость

равносильна

.
Легко видеть, что

делит

, а также

. Поэтому в виду взаимной простоты

, получаем, что

делит

.
Отсюда, в частности, следует, что

,

и поэтому

и

. То есть, для каждого фиксированного

количество решений конечно. Однако, описать их единой формулой - проблематично.
Вот для примера решения
![$[d,m,n]$ $[d,m,n]$](https://dxdy-04.korotkov.co.uk/f/f/c/c/fcc4df2387208315678113e53041af0e82.png)
для

:
Код:
[2, 3, 1]
[7, 9, 1]
[13, 3, 1]
[15, 4, 1]
[17, 4, 1]
[19, 5, 1]
[19, 16, 1]
[22, 137, 2]
[23, 10, 1]
[26, 55, 2]
[27, 2, 1]
[32, 7, 2]
[35, 9, 4]
[40, 9, 1]
[41, 9, 1]
[43, 185, 3]
[44, 41, 3]
[46, 5, 1]
[46, 559, 4]
[60, 253, 6]
[61, 39, 4]
[62, 297, 7]
[63, 193, 3]
[63, 44, 5]
[67, 17, 2]
[73, 53, 1]
[74, 33, 2]
[77, 4, 1]
[77, 67, 1]
[85, 16, 1]
[87, 8, 1]
[89, 144, 5]
[89, 151, 7]
[92, 19, 1]
[97, 68, 5]