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EXAMPLE
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We consider the case p = 1, n = 3. We have 5 solutions mod 3: (0,1,2), (0,2,1), (1,1,1), (1,2,0), (2,2,2).
With p = 2, n = 5, we have 12 solutions mod 5: (0,1,2), (0,1,3), (0,2,1), (0,2,4), (0,3,1), (0,3,4), (0,4,2), (0,4,3), (1,2,0), (1,3,0), (2,4,0), (3,4,0),
With p = 3, n = 7, we have 27 solutions mod 7: (0,1,3), (0,1,5), (0,1,6), (0,2,3), (0,2,5), (0,2,6), (0,3,1), (0,3,2), (0,3,4), (0,4,3), (0,4,5), (0,4,6), (0,5,1), (0,5,2), (0,5,4), (0,6,1), (0,6,2), (0,6,4), (1,3,0), (1,5,0), (1,6,0), (2,3,0), (2,5,0), (2,6,0), (3,4,0), (4,5,0), (4,6,0).
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