On the Parity of Some Quantities Related to the Euclidean Algorithm
1961
Proof of Theorem 1. By induction on n. If n =1 then xIjY, which implies x = 1 since (x, y) = 1. Hence x= 1 so that 0 y-xbl-r= 0 and y-f1b1-k y-bl(cx) r1= y, and y-f1b,-k
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