Volume 6, Number 3 June 2001 – October 2001
Pell’s Equation (I)
Kin Y. Li
Olympiad Corner
The 42nd International Mathematical Olympiad, Washington DC, USA, 8-9 July 2001
Problem 1. Let ABC be an acute-angled triangle with circumcentre O. Let P on BC be the foot of the altitude from A. Suppose that ∠BCA≥∠ABC+
!
30 . Prove that ∠CAB+∠COP<90!. Problem 2. Prove that
1 8 8 8 2 2 2+ + + + c + ab ≥ c ca b b bc a a
for all positive real numbers a, b and c.
Problem 3. Twenty-one girls and twenty-one boys took part in a mathematical contest.
• Each contestant solved at most six problems.
• For each girl and each boy, at least one problem was solved by both of them.
Prove that there was a problem that was solved by at least three girls and at least three boys.
(continued on page 4)
Editors: (CHEUNG Pak-Hong), Munsang College, HK (KO Tsz-Mei)
(LEUNG Tat-Wing), Applied Math. Dept., HKPU (LI Kin-Yin), Math. Dept., HKUST
(NG Keng-Po Roger), ITC HKPU
Artist: (YEUNG Sau-Ying Camille), MFA, CU
Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST
for general assistance.
On-line: http://www.math.ust.hk/mathematical_excalibur/
The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is
November 10, 2001.
For individual subscription for the next five issues for the 01-02 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI Department of Mathematics
The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
Let d be a positive integer that is not a square. The equation x2−dy2 =1 with variables x ,y over integers is called Pell’s equation. It was Euler who attributed the equation to John Pell (1611-1685), although Brahmagupta (7th century), Bhaskara (12th century) and Fermat had studied the equation in details earlier.
A solution (x,y) of Pell’s equation is called positive if both x and y are positive integers. Hence, positive solutions correspond to the lattice points in the first quadrant that lie on the hyperbola x2−dy2 =1. A positive solution )(x1,y1 is called the least
positive solution (or fundamental solution) if it satisfies x1<x and y < y1
for every other positive solution (x, y). (As the hyperbola x2−dy2=1 is strictly increasing in the first quadrant, the conditions for being least are the same as requiring x1+y1 d <x+
.) d y
Theorem. Pell’s equation x2−dy2 =1 has infinitely many positive solutions. If
) ,
(x1 y1 is the least positive solution,
then for n=1,2,3,..., define . ) ( 1 1 n n n y d x y d x + = +
The pairs (xn,yn) are all the positive solutions of the Pell’s equation. The
n
x ’s and y ’s are strictly increasing ton infinity and satisfy the recurrence relations xn+2=2x1xn+1−xn and yn+2
. 2x1yn 1−yn
= +
We will comment on the proof. The least positive solution is obtained by writing d as a simple continued fraction. It turns out
" 1 1 1 2 1 0 + + + = a a a d
,
where a0=[ d] and a1,a2,... is a periodic positive integer sequence. The continued fraction will be denoted by... , , , 1 2 0 a a a . The k-th convergent of ... , , , 1 2 0 a a a is the number k k q p = k a a a a0, 1, 2,..., with p ,k qk relatively prime. Let a1,a2,...,am be the period
for d . The least positive solution of Pell’s equation turns out to be
= − − − − odd is if ) , ( even is if ) , ( ) , ( 1 2 1 2 1 1 1 1 m q p m q p y x m m m m .
For example, 3 = 1,2 ,2,1 ,2,... and so m = 2, then 1 2 1 , 1 = . We check 2 2 1 3
2 − ⋅ = 1 and clearly, (2 ,1) is the least positive solution of x2 −3y2 =1. Next, 2 = 1,2,2 ,... and so m=1, then . 2 3 2 , 1 = We check 32−2⋅22 = 1 and again clearly (3,2) is the least
positive solution of x2−2y2 =1. Next, if there is a positive solution (x, y) such that xn+ yn d <x+y d <xn+1 d yn 1+ + , then consider u+v d = ) /( ) (x+ y d xn+yn d . We will get u + v d <x1+y1 d and u−v d = ) /( ) (x−y d xn −yn d so that u2 − 2 dv = (u−v d)(u+v d) = 1, con-tradicting )(x1,y1 being the least
positive solution.
To obtain the recurrence relations, note that
Mathematical Excalibur, Vol. 6, No. 3, Jun 01- Oct 01 Page 2 d y x dy x d y x1 1 )2 12 12 2 1 1 ( + = + + d y x x12 1 2 1 1 2 − + = 1 ) ( 2 1 1+ 1 − = x x y d . So d y xn+2+ n+2 n d y x d y x ) ( ) ( 1+ 1 2 1+ 1 = n n x y d d y x x ( ) ( ) 2 1 1+ 1 1− 1+ 1 = + . ) 2 ( 2x1xn 1−xn + x1yn 1−yn d = + +
The related equation x2 −dy2 =−1 may not have a solution, for example, x2 −
1 3y2 =− cannot hold as x2 −3y2 ≡ 1 2 2+ ≠− y x (mod 4). However, if d is a prime and d ≡1 (mod 4), then a theorem of Lagrange asserts that it will have a solution. In general, if x2−dy2
1
−
= has a least positive solution )
,
(x1 y1 , then all its positive solutions
are pairs (x, y), where x + y d =
1 2 1
1 )
(x +y d n− for some positive integer n.
In passing, we remark that some k-th convergent numbers are special. If the length m of the period for d is even, then x2 −dy2 =1 has (xn,yn)=
) ,
(pnm−1 qnm−1 as all its positive
solutions, but x2−dy2 =−1 has no integer solution. If m is odd, then
1
2
2− =
dy
x has (pjm−1 ,yjm−1) with j even as all its positive solutions and
1
2
2− =−
dy
x has (pjm−1 ,qjm−1) with j odd as all its positive solutions.
Example 1. Prove that there are infinitely many triples of consecutive integers each of which is a sum of two squares.
Solution. The first such triple is 8=22 , 1 3 10 , 0 3 9 , 22 = 2+ 2 = 2+ 2 + which
suggests we consider triples x2−1, x2, .
1
2+
x Since x2− 2y2 = 1 has infinitely many positive solutions (x, y), we see that x2−1=y2+y2, x =2 x +2
2
0 and x2+1 satisfy the requirement and there are infinitely many such triples.
Example 2. Find all triangles whose sides are consecutive integers and areas are also integers.
Solution. Let the sides be z – 1, z, z + 1.
Then the semiperimeter 2 3z s= and the area is 4 ) 4 ( 3 2− = z z A . If A is an
integer, then z cannot be odd, say z = 2x, and 4z2 − = 3ω2. So 4x2−4=3ω2, which implies ω is even, say ω=2 y. Then x2−3y2 =1, which has (x1,y1)= (2, 1) as the least positive solution. So all positive solutions are (xn,yn), where xn
. ) 3 2 ( 3 n n y = + + Now xn −yn 3= n ) 3 2 ( − . Hence, 2 ) 3 2 ( ) 3 2 ( n n n x = + + − and 3 2 ) 3 2 ( ) 3 2 ( n n n y = + − − .
The sides of the triangles are 2xn −1, 1
2 ,
2xn xn + and the areas are A = .
3xnyn
Example 3. Find all positive integers k, m such that k < m and
. ) 2 ( ) 1 ( 2 1+ +#+k= k+ + k+ +#+m Solution. Adding 1+2+#+k to both sides, we get 2k(k+1)=m(m+1), which can be rewritten as (2m+1)2−2(2k+1)2 = .−1 Now the equation x2 −2y2=−1 has (1,1) as its least positive solution. So its positive solutions are pairs
1 2 ) 2 1 ( 2 = + − + n n n y x . Then 2 ) 2 1 ( ) 2 1 ( + 2 −1+ − 2 −1 = n n n x and 2 2 ) 2 1 ( ) 2 1 ( + 2 −1− − 2 −1 = n n n y .
Since x2−2y2 =−1 implies x is odd, so x is of the form 2m+1. Then y2 =2m2+
1
+
m implies y is odd, so y is of the form
. 1 2k+ Then − − = 2 1 , 2 1 ) , (k m yn xn
with n = 2, 3, 4, … are all the solutions.
Example 4. Prove that there are infinitely many positive integers n such that n2+1 divides n!.
Solution. The equation x2 −5y2 =−1 has (2,1) as the least positive solution. So it has infinitely many positive solutions. Consider those solutions with
. 5 > y Then 5<y<2y≤x as 4 y2≤ . 1 5y2− =x2 So 2(x2+1)=5⋅y⋅2y
divides x!, which is more than we want.
Example 5. For the sequence a =n
2+ + 2 ) 1 (n
n , prove that there are
infinitely many n’s such that an−an+1> 1 and an+1−an =1.
Solution. First consider the case n2+ ,
) 1
(n+ 2= y2 which can be rewritten as . 1 2 ) 1 2 ( n+ 2− y2 =− As in example 3 above, x2−2y2=−1 has infinitely many positive solutions and each x is odd, say x=2n+1 for some n. For these n’s, an =y and an−1 = − 2+ 2 ) 1 (n n = y2 −4n . The equation y2 =n2+(n+1)2 implies n > 2 and an−1≤ y2−4n <y−1=an−1. So an − an−1>1 for these n’s.
Also, for these n’s, an+1 =
. 4 4 ) 2 ( ) 1 ( 2 2 2 + + = n+ + n+ y n As n < y < 2n + 1, we easily get y + 1 < . 2 4 4 2+ + < + y n y So an+1−an = (y + 1) – y = 1.
Example 6. (American Math Monthly E2606, proposed by R.S. Luthar) Show that there are infinitely many integers n such that 2n + 1 and 3n + 1 are perfect squares, and that such n must be multiples of 40.
Solution. Consider 2n + 1 = u and 3n2
2
1=v
+ . On one hand, u2+v2 ≡2 (mod 5) implies u , 2 v2 ≡ 1 (mod 5), which means n is a multiple of 5.
On the other hand, we have 3u2 −2v2 = 1. Setting u = x + 2y and v = x + 3y, the equation becomes x2−6y2 =1. It has infinitely many positive solutions. Since
, 1 2
3u2 − v2 = u is odd, say u = 2k + 1. Then n = 2k + 2k is even. Since 3n + 12 = v , so v is odd, say v = 4m2 ±1. Then 3n = 16m2±8m, which implies n is also a multiple of 8.
Mathematical Excalibur, Vol. 6, No. 3, Jun 01- Oct 01 Page 3
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, Hong Kong University of Science & Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is November 10, 2001.
Problem 131. Find the greatest common divisor (or highest common factor) of the numbers nn −n for n = 3, 5, 7, … . Problem 132. Points D, E, F are chosen on sides AB, BC, CA of ABC∆ , respectively, so that DE = BE and FE = CE. Prove that the center of the circumcircle of ∆ADF lies on the angle bisector of ∠DEF. (Source: 1989 USSR Math Olympiad)
Problem 133. (a) Are there real numbers a and b such that a+b is rational and an +bn is irrational for every integer n≥2? (b) Are there real numbers a and b such that a+b is irrational and an +bn is rational for every integer n≥2? (Source: 1989 USSR Math Olympiad)
Problem 134. Ivan and Peter alternatively write down 0 or 1 from left to right until each of them has written 2001 digits. Peter is a winner if the number, interpreted as in base 2, is not the sum of two perfect squares. Prove that Peter has a winning strategy. (Source: 2001 Bulgarian Winter Math Competition)
Problem 135. Show that for n≥2, if , 0 ..., , , 2 1 a an > a then ≥ + + +1)( 1) ( 1) ( 23 3 3 1 a an a # ). 1 ( ) 1 )( 1 ( 1 2 3 2 2 2 2 1a + a a + a a + a # n
(Source: 7th Czech-Slovak-Polish Match)
*****************
Solutions
*****************
Problem 126. Prove that every integer
can be expressed in the form x2+y2−
,
5z2 where x, y, z are integers.
Solution. CHAN Kin Hang (CUHK, Math Major, Year 1), CHENG Kei Tsi Daniel (La Salle College, Form 7), CHENG Man Chuen (CUHK, Math Major, Year 1), CHUNG Tat Chi (Queen Elizabeth School, Form 5), FOK Chi Kwong (Yuen Long Merchants Association Secondary School, Form 5), IP Ivan (St. Joseph’s College, Form 6), KOO Koopa (Boston College, Sophomore), LAM Shek Ming Sherman (La Salle College, Form 6), LAU Wai Shun (Tsuen Wan Public Ho Chuen Yiu Memorial College, Form 6), LEE Kevin (La Salle College, Form 6), 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), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6), YEUNG Kai Sing (La Salle College, Form 5) and YUNG Po Lam (CUHK, Math Major, Year 2). For n odd, say n = 2k + 1, we have
2 2 2 5 ) 1 ( ) 2 ( k + k+ − k = 2k + 1 = n. For n even, say n = 2k, we have (2k−1)2+
2 2 ) 1 ( 5 ) 2 (k− − k− = 2k = n.
Problem 127. For positive real numbers a, b, c with a + b + c = abc, show that 2 3 1 1 1 1 1 1 2 2 2 + + + + ≤ +a b c ,
and determine when equality occurs. (Source: 1998 South Korean Math Olympiad)
Solution. CHAN Kin Hang (CUHK, Math Major, Year 1), CHENG Kei Tsi Daniel (La Salle College, Form 7), KOO Koopa (Boston College, Sophomore), LEE Kevin (La Salle College, Form 6) and NG Ka Chun (Queen Elizabeth School, Form 7).
Let A = tan−1a, B = tan−1b, C = tan−1c. Since a, b, c > 0, we have 0 < A, B, C <
2
π
. Now a + b + c = abc is the same as
tan A + tan B + tan C = tan A tan B tan C. Then tan C = B A B A tan tan 1 ) tan (tan − + − = tan(π– A – B) which implies A + B + C = π. In terms of A, B, C the inequality to be proved is
cos A + cos B + cos C 2 3
≤ , which follows by applying Jensen’s inequality
to f(x) = cos x on ). 2 , 0 ( π
Other commended solvers: CHENG Man Chuen (CUHK, Math Major, Year 1), IP Ivan (St. Joseph’s College, Form 6), LAM Shek Ming Sherman (La Salle College, Form 6), LEUNG Wai Ying (Queen Elizabeth School, Form 7), MAN Chi Wai (HKSYC&IA Wong Tai Shan Memorial College), TSUI Ka Ho (Hoi Ping Chamber of Commerce Secondary School, Form 7), WONG Wing Hong (La Salle College, Form 4) and YEUNG Kai Sing (La Salle College, Form 5).
Problem 128. Let M be a point on segment AB. Let AMCD, BEHM be squares on the same side of AB. Let the circumcircles of these squares intersect at M and N. Show that B, N, C are collinear and H is the orthocenter of
. ABC
∆ (Source: 1979 Henan Province Math Competition)
Solution. LEUNG Wai Ying (Queen Elizabeth School, Form 7), MAN Chi Wai (HKSYC&IA Wong Tai Shan Memorial College) and YUNG Po Lam (CUHK, Math Major, Year 2).
Since ∠BNM = ∠BHM = 45 =! CDM
∠ = ∠CDM , it follows B, N, C are collinear. Next, CH⊥AB. Also,
ME
BH⊥ and ME AC imply BH ⊥ AC. So H is the orthocenter of ∆ABC. Other commended solvers: CHAN Kin Hang (CUHK, Math Major, Year 1), CHENG Kei Tsi Daniel (La Salle College, Form 7), CHENG Man Chuen (CUHK, Math Major, Year 1), CHUNG Tat Chi (Queen Elizabeth School, Form 5), IP Ivan (St. Joseph’s College, Form 6), KWOK Sze Ming (Queen Elizabeth School, Form 6), LAM Shek Ming Sherman (La Salle College, Form 6), Lee Kevin (La Salle College, Form 6), NG Ka Chun (Queen Elizabeth School, Form 7), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6), WONG Wing Hong (La Salle College, Form 4) and YEUNG Kai Sing (La Salle College, Form 5).
Problem 129. If f(x) is a polynomial of degree 2m + 1 with integral coefficients for which there are 2m + 1 integers
1 2 2
1,k , ,k m+
k $ such that f(ki)=1 for i = 1, 2, …, 2m + 1, prove that f(x) is not the product of two nonconstant poly-nomials with integral coefficients. Solution. CHAN Kin Hang (CUHK, Math Major, Year 1), CHENG Kei Tsi Daniel (La Salle College, Form 7), CHENG Man Chuen (CUHK, Math Major, Year 1), IP Ivan (St. Joseph’s College, Form 6), KOO Koopa (Boston College, Sophomore), LAM Shek Ming Sherman (La Salle College, Form 6),
Mathematical Excalibur, Vol. 6, No. 3, Jun 01- Oct 01 Page 4 LEE Kevin (La Salle College, Form 6),
LEUNG Wai Ying (Queen Elizabeth School, Form 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)=
) ( ) (ki Q ki
P and P(ki), Q(ki) are integers, so either both are 1 or both are –1. As there are 2m + 1k ’s, eitheri
) ( ) (ki Q ki P = = 1 for at least m + 1k ’si or 1P(ki)=Q(ki)=− for at least m + 1k ’s. Since deg f = 2m + 1, one of degi 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).
Let P = 3 1 ,
2 . Suppose lattice points
) , ( ), ,
(x0 y0 x1 y1 are the same distance
from P. Then
(
)
2 0 2 0 3 1 2 − + − y x =(
)
. 3 1 2 2 1 2 1 − + − y x Moving the xterms to the left, the y terms to the right and factoring, we get
(x0−x1)
(
x0+x1−2 2)
= . 3 2 ) ( 0 1 0 1 + − −y y y yAs the right side is rational and 2 is irrational, we must have x0 =x1. Then the left side is 0, which forces y1= y0 since y1+ y0 is integer. So the lattice
points are the same.
Now consider the circle with center at 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).
Olympiad Corner
(continued from page 1)
Problem 4. Let n be an odd integer greater than 1, let k1,k2 ,...,kn be given integers. For each of the n! permutations
) ..., , , (a1 a2 an a= of 1, 2, …, n, let . ) ( 1
∑
= = n i i ia k a SProve that there are two permutations b and c, b ≠ c, such that n! is a divisor of S(b) – S(c).
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=60! and that AB + BP = AQ + QB.
What are the possible angles of triangle ABC?
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.
Pell’s Equation (I)
(continued from page 2)
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 implies a is odd.
Setting 1x=3b+ and y=3(b−1)/2 2 / ) 1 ( 5a− , we get 15y2 =3b5a =3b(3b+ 2) = (3b+1)2−1= x2 −1. So (x, y) is a positive solution of x2−15y2 =1. The least positive solution is (4 ,1). Then (x, y) = (xn,yn) for some positive integer n, where xn +yn 15=(4+ 15)n. After examining the first few y ’s, wen observe that y3k are the only terms that are divisible by 3. However, they also seem to be divisible by 7, hence cannot be of the form 3c5d.
To confirm this, we use the recurrence relations on y . Since n y1 =1, y2 =8 and yn+2 =8yn+1−yn, taking y (modn 3), we get the sequence 1, 2, 0, 1, 2, 0… and taking y (mod 7), we get 1, 1, 0, -1,n -1, 0, 1, 1, 0, -1, -1, 0, ….
Therefore, no y= ynis of the form 3c5d and 1a, b> cannot be solution to
b a
3 5 − = 2.
Example 8. Show that the equation a2
4 3
c b =
+ has infinitely many solutions. Solution. We will use the identity
+ 3 1 , 2 ) 1 ( 2 2 3 3 + = + +# n n n
which is a standard exercise of mathematical induction. From the
identity, we get + = − 2 ) 1 ( 3 2 n n n 2 2 ) 1 ( n n+
for n > 1. All we need to do
now is to show there are infinitely many positive integers n such that n(n + 1)/2 =
2
k for some positive integers k. Then (a, b, c) = ((n – 1)n/2, n, k) solves the problem.
Now n(n + 1)/2 = k can be rewritten as2 . 1 ) 2 ( 2 ) 1 2 ( n+ 2− k 2 = We know x2 − 1
2y2 = has infinitely many positive solutions. For any such (x ,y), clearly x is odd, say x=2m+1. They y2 =2m2
m 2
+ implies y is even. So any such (x, y) is of the form (2n+1, 2k). Therefore, there are infinitely many such n.
