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... not. The sum of all terms of the latter type we recognize as 2y(p?r 1 -p^ p%h). Each of the remaining terms is of the form y(pnd), where d properly divides p?2 p%h. Moreover, in each case, d has lower exponent than that of N/p^K This observation leads us to a proof by induction on the exponent of N ...
... not. The sum of all terms of the latter type we recognize as 2y(p?r 1 -p^ p%h). Each of the remaining terms is of the form y(pnd), where d properly divides p?2 p%h. Moreover, in each case, d has lower exponent than that of N/p^K This observation leads us to a proof by induction on the exponent of N ...