Mathematical Excalibur, Vol. 6, No. 3, Jun 01- Oct 01
7), MAN Chi Wai (HKSYC&IA Wong Tai Shan Memorial College), YEUNG Kai Sing (La Salle College, Form 5) and YUNG Po Lam (CUHK, Math Major, Year 2). Suppose f is the product of two non-constant polynomials with integral co-efficients, say f = PQ. Since 1 = f ( ki ) = P (ki ) Q (ki ) and P ( ki ), Q ( ki ) are integers, so either both are 1 or both are –1. As there are 2m + 1 k i ’s, either P (k i ) = Q (k i ) = 1 for at least m + 1 k i ’s or P ( k i ) = Q ( k i ) = −1 for at least m + 1 k i ’s. Since deg f = 2m + 1, one of deg P or deg Q is at most m. This forces P or Q to be a constant polynomial, a contradiction. Other commended solvers: NG Cheuk Chi (Tsuen Wan Public Ho Chuen Yiu Memorial College) and NG Ka Chun (Queen Elizabeth School, Form 7). Problem 130. Prove that for each positive integer n, there exists a circle in the xy-plane which contains exactly n lattice points in its interior, where a lattice point is a point with integral coordinates. (Source: H. Steinhaus, Zadanie 498, Matematyka 10 (1957), p. 58) Solution. CHENG Man Chuen (CUHK, Math Major, Year 1) and IP Ivan (St. Joseph’s College, Form 6). 1⎞ ⎛ Let P = ⎜ 2 , ⎟ . 3⎠ ⎝
Suppose lattice
points ( x0 , y 0 ), ( x1 , y1 ) are the same distance
(x (x
0
− 2
1− 2
from
)
2
)
2
P.
1⎞ ⎛ + ⎜ y0 − ⎟ 3⎠ ⎝
Then
2
Page 4
P and radius r. As r increases from 0 to infinity, the number of lattice points inside the circle increase from 0 to infinity. As the last paragraph shows, the increase cannot jump by 2 or more. So the statement is true. Other commended solvers: CHENG Kei Tsi Daniel (La Salle College, Form 7), KOO Koopa (Boston College, Sophomore), LEUNG Wai Ying (Queen Elizabeth School, Form 7), MAN Chi Wai (HKSYC&IA Wong Tai Shan Memorial College), NG Ka Chun (Queen Elizabeth School, Form 7) and YEUNG Kai Sing (La Salle College, Form 4).
implies a is odd. x = 3b + 1
Setting
2
y = 3( b−1) / 2
5( a −1) / 2 , we get 15 y 2 = 3b5a = 3b (3b +
2) = ( 3b + 1) 2 − 1 = x 2 − 1. So (x, y) is a positive solution of x 2 − 15 y 2 = 1. The least positive solution is ( 4, 1) . Then (x, y) = ( x n , y n ) for some positive integer n, where x n + y n 15 = ( 4 + 15 ) n .
After
examining the first few y n ’s, we observe that y 3k are the only terms that are divisible by 3. However, they also seem to be divisible by 7, hence cannot be of the form 3c 5d .
Olympiad Corner (continued from page 1)
Problem 4. Let n be an odd integer greater than 1, let k1 , k 2 , ..., k n be given integers. For each of the n! permutations a = ( a1 , a 2 , ..., a n ) of 1, 2, …, n, let n
S ( a ) = ∑ ki ai . i =1
To confirm this, we use the recurrence relations on y n . Since y1 = 1, y 2 = 8 and y n + 2 = 8 y n +1 − y n , taking y n (mod 3), we get the sequence 1, 2, 0, 1, 2, 0… and taking y n (mod 7), we get 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, …. Therefore, no y = y n is of the form 3c 5d and a, b > 1 cannot be solution to 5a − 3b = 2.
Prove that there are two permutations b and c, b ≠ c, such that n! is a divisor of S(b) – S(c).
Example 8. Show that the equation a 2 + b 3 = c 4 has infinitely many solutions.
Problem 5. In a triangle ABC, let AP bisect ∠BAC , with P on BC, and let BQ bisect ∠ABC , with Q on CA. It is known that ∠BAC = 60o and that AB + BP = AQ + QB. What are the possible angles of triangle ABC?
⎛ n( n + 1) ⎞ 13 + 2 3 + L + n 3 = ⎜ ⎟ , 2 ⎠ ⎝
Solution. We will use the identity 2
which
is
a
Problem 6. Let a, b, c, d be integers with a > b > c > d > 0. Suppose that ac + bd = (b + d + a – c)(b + d – a + c). Prove that ab + cd is not prime.
standard
mathematical induction.
exercise
of
From the 2
identity,
we
=
1⎞ ⎛ + ⎜ y1 − ⎟ . Moving the x 3⎠ ⎝
and
get
⎛ ( n − 1)n ⎞ 3 ⎜ ⎟ +n = 2 ⎝ ⎠
2
⎛ n( n + 1) ⎞ ⎜ ⎟ for n > 1. All we need to do ⎝ 2 ⎠
now is to show there are infinitely many
terms to the left, the y terms to the right
positive integers n such that n(n + 1)/2 =
and factoring, we get
k 2 for some positive integers k. Then (a,
(
( x0 − x1 ) x0 + x1 − 2 2
)
Pell’s Equation ( I ) (continued from page 2)
2⎞ ⎛ = ( y 0 − y1 )⎜ y 0 + y1 − ⎟ . 3⎠ ⎝
As the right side is rational and 2 is irrational, we must have x0 = x1. Then the left side is 0, which forces y1 = y 0 since y1 + y 0 is integer. So the lattice points are the same. Now consider the circle with center at
Example 7. Prove that the only positive integral solution of 5a − 3b = 2 is a = b = 1. Solution. Clearly, if a or b is 1, then the other one is 1, too. Suppose (a, b) is a solution with both a, b > 1. Considering (mod 4), we have 1 – ( −1) b ≡ 2 (mod 4), which implies b is odd. Considering (mod 3), we have ( −1) a ≡ 2 (mod 3), which
b, c) = ((n – 1)n/2, n, k) solves the problem. Now n(n + 1)/2 = k 2 can be rewritten as ( 2n + 1) 2 − 2( 2k ) 2 = 1. We know x 2 − 2 y 2 = 1 has infinitely many positive
solutions. For any such ( x, y ), clearly x is odd, say x = 2m + 1. They y 2 = 2m 2 +2m implies y is even. So any such (x, y)
is of the form ( 2n + 1, 2k). Therefore, there are infinitely many such n.