- #1
Ben2
- 37
- 9
From Gauss, Disquisitiones Arithmeticae p. 10: Given A(1) = a(1) = [a(1)], A(2) = a(2)A(1) + 1 = [a(1), a(2)]and for n>= 3, A(n) = a(n)A(n-1) + A(n-2) = [a(1), a(2), ... , a(n)], prove that [a(1), a(2), ... , a(n)]*[a(2), a(3), ... , a(n-1)] - [a(1), a(2), ... , a(n-1)]*[a(2), a(3), ... , a(n)] = (-1)^n, where "^" indicates "to the power", and that
[a(1), a(2), ... , a(n)] = [a(n), a(n-1), ... , a(1)].
This stems from an interest in congruences, coming in turn from Nishiyama's online article on the strobe effect--a physics question. Thanks to all who comment.
[a(1), a(2), ... , a(n)] = [a(n), a(n-1), ... , a(1)].
This stems from an interest in congruences, coming in turn from Nishiyama's online article on the strobe effect--a physics question. Thanks to all who comment.