2.
74
2. 2.
NUMBER THEORY
{:}
{:}
cp( )
cp( )
cp( ) -
hence d<n gcd(d, n ) = l
On the other hand,
d = ncp(n) 2 .
d = L (n + d) = ncp(n) + L d = = 3ncp(n) = ncp(n) + ncp(n) . 2 2 Therefore 3ncp(n) = ncp(n) 2 + 2 = 2ncp(n). (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 61, Problem 4574) 42. We proceed by induction. For n = 3 the claim is true. Assume that the hypothesis holds for n - 1. Let 1 < k < n! and let kl , q be the quotient and the remainder when k is divided by n. Hence k = k n + q, 0 � q < n and 0 � kl < � < n n!- = (n - I)!. n From the inductive hypothesis, there are integers di < d� < . . . < d�, s � i = 1,2, . . . , s and kl = di + d� + . . . + d�. Hence n - 1, such that. .dil(n I)!, k = ndi + nd� + . + nd� + q. If q = 0, then k = d1 + d2 + . . . + ds, where di = ndi, i = 1,2, . . . , s, are distinct divisors of n!. If q f. 0, then k = d1 + rh + . . . + ds+1 , where di = ndL iI, 2, . . . , s, and ds+1 = q. It is clear that dil n !, i = 1, 2, . . , s and ds+1 I n !, since q < n. On the other hand, dS+1 < d1 < d2 < . . < ds, because dS+1 = q < n � ndi = d1 . Therefore k can be written as a sum of at most n distinct divisors of n!, as claimed. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1983), pp. 88, Problem C4:10) L
d<n gcd( d , n ) = l
n < d <2n gcd( d , n ) =l
d<n gcd(d , n ) = l
s
i
.
.
75
43. If n � 992, take the set with all 992 odd numbers from {I, 2, . . . , 1984}.
Note that
gcd(n, d) = 1 gcd(n, n - d) = 1 gcd(n, n + d) = 1 (1) Let d1 , d , . . . ,d be the numbers less than n and relatively prime with n. From (1) we deduce2 that n d1 + d n = n d2 + d n l = n
SOLUTIONS
Its complementary set has only even numbers, any two of them not being relatively Let be the complementary set of a subset with elements prime. Hence � of the set Define D E If n D 0, then C U D has elements but u D C which is false. Hence n D f. 0, so there is an element E n D. It follows that E and + E and since and + are relatively prime we are done. Revista Matematica Timi§oara (RMT) , No. pp. Problem
n 991. c 991 {I, 2, . . . , 1984}. = {c + 11 c C}. C = C {I, 2, . . . , 1985}, 2 · 993 = 1986 C a C a C a a 1 a 1 C (Titu Andreescu, 1(1984), 102, C4:7) 44. Let a be the greatest odd integer such that a2 < n, hence n < (a + 2) 2 . If a � 7, then a - 4, a - 2, a are odd integers which divide n. Note that any two of these numbers are relatively prime, so (a - 4)(a - 2)a divides n. It follows that (a -4)(a - 2)a < (a + 2) 2 so a3 -6a2 + 8a < a2 + 4a + 4. Then a3 - 7a2 + 4a - 4 � 0 or a2 (a - 7) + 4(a - 1) � O. This is false, because a � 7, hence a = 1,3 or 5. If a = 1, then 1 2 � n < 32 , so n E {I, 2, . . . , 8}. If a = 3, then 32 � n < 52 and 1 . 31 n , so n E {9, 12, 15, 18,21, 24}. If a = 5, then 52 � n < 72 and 1 . 3 . 51 n so n E {30,45}. Therefore n E {2,3, 4,5,6, 7,8,9, 12,15, 18,21,24,30,35}. (Dorin Andrica and Adrian P. Ghioca, Romanian Winter Camp 1984; Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 78, Problem T. 2 1) 45. We prove by induction that
u� - 2v� = 1, n � 1.
(1)
n = 1 the claim is true. Assuming that the equality is true for some n, we U�+l - 2V�+1 = (3un + 4Vn)2 - 2(2un + 3Vn )2 = u� - 2v� = 1 hence (1) is true for all n � 1. We prove now that (2) 2x� - Y� = 1, n � 1 Indeed, 2x� - Y� = 2(un + Vn )2 - (un + 2vn)2 = u� - 2v� = 1,
For have
as
claimed. It follows that
(Xn V2 - Yn) (Xn V2 + Yn) = 1, n � 1. Notice that xn ..,fi + Yn > 1 so 0 < xn V2 -Yn < 1, n � 1. Hence Yn = [Xn V2] , as claimed.