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inductive step, use antanairesis and the deﬁning recurrence,
( Fn +2 , Fn +1 ) = ( Fn +1 − Fn , Fn +1 )
= ( Fn , Fn +1 ) = 1.
Here is a proof that is a variation of the same idea. Let n ≥ 1 be the
smallest integer for which Fn +1 and Fn have gcd d > 1. We note that
n > 1 because ( F2 , F1 ) = (1, 1) = 1, and so n − 1 ≥ 1. But if d is a
common divisor of Fn +1 and Fn , then d divides Fn −1 = Fn +1 − Fn , so
that ( Fn , Fn −1 ) = 1. This contradicts n being the smallest index for which
( Fn +1 , Fn ) = 1.
(i) Show that if d is the greatest common divisor of a1 , a2 , . . . , an ,
then d = ti ai , where ti is in Z for all i with 1 ≤ i ≤ n .
Solution. The set I of all linear combinations ti ai of a1 , a2 , . . . ,
an , where ti is in Z for 1 ≤ i ≤ n , satisﬁes the conditions of Corollary 1.37. If d is the smallest positive element in I , then the proof
of Theorem 1.35 can be modiﬁed to show that d is the gcd.
(ii) Prove that if c is a common divisor of a1 , a2 , . . . , an , then c | d .
Solution. The proof of Corollary 1.40 generalizes easily.
1.66 (i) Show that (a , b, c), the gcd of a , b, c, is equal to (a , (b, c)).
Solution. It sufﬁces to prove that any common divisor of a , b, c
is a common divisor of a and (b, c), and conversely. But each of
these statements is easy to prove. (ii) Compute (120, 168, 328).
(120, 168, 328) = (120, (328, 168)) = (120, 8) = 8
1.67 (i) Consider a complex number z = q + i p, where q > p are positive
integers. Prove that
(q 2 − p2 , 2q p , q 2 + p2 )
is a Pythagorean triple by showing that |z 2 | = |z |2 .
Solution. If z = q + i p, then |z 2 | = |z |2 , by part (i). Now
z 2 = (q 2 − p2 ) + i 2q p , so that |z 2 | = (q 2 − p2 )2 + (2q p )2 . On
the other hand, |z |2 = (q 2 + p2 )2 . Thus, if we deﬁne a = q 2 − p2 ,
b = 2q p , and c = q 2 + p 2 , then a 2 + b2 = c2 and (a , b, c) is a
Pythagorean triple. (ii) Show that the Pythagorean triple (9, 12, 15) (which is not primitive) is not of the type given in part (i). ...
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- Fall '11
- Greatest common divisor, Euclidean algorithm, FN, common divisor