Problem 102, January 2009
Find all real solutions to the following system

 1 − x2
1 − y2

1 − z2
of equations:
= y
= z .
= x
Problem 102, January 2009 - Solution
If any of x, y, z is nonnegative, we may assume x ≥ 0.
Then from 1 − z 2 = x we deduce that 0 ≤ x ≤ 1 and thus, x2 ≤ 1.
Then from 1 − x2 = y we deduce that 0 ≤ y ≤ 1 and thus, y 2 ≤ 1.
Then from 1 − y 2 = z we deduce that z ≥ 0.
Hence, x, y, z are all nonnegative. Without loss of generality, assume x = max{x, y, z}.
Then z 2 = 1 − x ≤ 1 − z = y 2 ⇒ z ≤ y and y = 1 − x2 ≤ 1 − y 2 = z ⇒ y ≤ z.
Thus, y = z and it then follows that x = y = z.
If x < 0, y < 0, and z <
becomes
 2
 x1 − 1
y2 − 1
 12
z1 − 1
0, let x1 = −x, y1 = −y, and z1 = −z. Then the given system
= y1
= z1
= x1
where x1 > 0, y1 > 0, z1 > 0.
Without loss of generality, assume x1 = max{x1 , y1 , z1 }.
Now z12 = 1 + x1 ≥ 1 + z1 = y12 ⇒ z1 ≥ y1
y1 = x21 − 1 ≥ y12 − 1 = z1 ⇒ y1 ≥ z1
Thus, y1 = z1 and it then follows that x1 = y1 = z1 . Hence, x = y = z. Therefor,
x = y = z in all cases.
−1 ±
Solving 1 − x = x we get x = y = z =
2
2
√
5
.
Problem 103, February 2009
Let m, n be positive integers such that (m, n) = 1. Determine the possible values of
gcd(7m + 5m , 7n + 5n ).
Problem 103, February 2009 - Solution
Let fm = 5m + 7m , fn = 5n + 7n and let d = (fm , fn ).
First, if m = n, then (m, n) = 1 ⇒ m = n = 1 ⇒ d = (12, 12) = 12.
Without loss of generality, we may assume that m < n.
If n = 2m, then 1 = (m, n) = m ⇒ n = 2. Hence, d = (f1 , f2 ) = (12, 74) = 2. Now,
suppose n > 2m. Then from fm · fn−m = (5m + 7m )(5n−m + 7n−m ) = fn + 5m 7m fn−2m we
obtain fn = fm fn−m − 5m 7m fn−2m .
Hence,
d = (fm , fn ) = (fm , fn−2m )
(1)
Similarly, if m < n < 2m, then from fm · fn−m = fn + 5n−m 7n−m f2m−n we obtain
fn = fm fn−m − 5n−m 7n−m f2m−n .
Hence,
d = (fm , fn ) = (fm , f2m−n )
(2)
By iterating (1) and (2) and using the Euclidean Algorithm, we conclude that
d = (f1 , f1 ) = 12 if m + n is even; that is, if m and n are both odd since m and n clearly
can not both be even. If m + n is odd, then we would have d = (f1 , f2 ) = (12, 74) = 2.
Note that these answers are consistent with the special cases when m = n = 1 or m = 1,
n = 2.
Problem 104, March 2009
The sequence {an }∞
n=1 is defined by a1 = 1 and an+1 =
ba2n c = n for all n ≥ 4.
an
n
+
for all n ≥ 1. Prove that
n
an
Problem 104, March 2009 - Solution
2 3
13
Direct computations show that a2 = 2, a3 = 2, and a4 = + = .
3 2
6
169
2
2
c = 4. Suppose ban c = n for some n ≥ 4.
Hence, ba4 c = b
36
Then n ≤ a2n < n + 1
(∗)
a2n n2
+ . Then ba2n+1 c = n+1 ⇔ bS+2c = n+1 ⇔ bSc = n−1 ⇔ n−1 ≤ S < n.
n2 a2n
2
n
(n + 1)2 n2
1
1
Using (*) we have S <
+ 2 = 1+
+1 < 1+
+ 1 < e + 1 < 4 ≤ n.
n2
n
n
n
On the other hand,
Let S =
a2n n2
+ 2 ≥ n − 1 ⇔ a4n + n4 − n2 (n − 1)a2n ≥ 0
2
n
an
4
2 2
3
⇔ an + n an + n (n + 1 − a2n ) − n3 ≥ 0
n−1≤S =
⇔ a4n + n2 (a2n − n) + n3 (n + 1 − a2n ) ≥ 0
which is true by (*).
Hence, ba2n+1 c = n + 1 and the induction is complete.
Problem 105, April 2009
Let f (n) = n2 − 19n + 99 where n ∈ N. Determine all values of n for which f (n) is a perfect
square.
Problem 105, April 2009 - Solution
Let f (n) = n2 − 19n + 99 = (n − 9)(n − 10) + 9. Claim that (n − 10)2 < f (n) < (n − 9)2
for all n > 18 (*)
Since f (n) − (n − 10)2 = (n2 − 19n + 99) − (n2 − 20n + 100) = n − 1 > 0 for all n > 1 and
(n − 9)2 − f (n) = (n2 − 18n + 81) − (n2 − 19n + 99) = n − 18 > 0 for all n > 18 (*) follows.
Thus, f (n) can’t be a square if n > 18. Next, note that f (19−n) = (10−n)(9−n)+9 = f (n)
and so it suffices to check f (1), f (2), · · · f (9):
f (1) = f (18) = 81 = 92 , f (2) = f (17) = 65, f (3) = f (16) = 51,
f (4) = f (15) = 39, f (5) = f (14) = 29, f (6) = f (13) = 21,
f (7) = f (12) = 15, f (8) = f (11) = 11, f (9) = f (10) = 9 = 32
Hence the only values of n for which f (n) is a square are n = 1, 9, 10, 18.
Problem 106, May 2009
Solve the equation below:
(2 cos x − 1)(2 cos 2x − 1)(2 cos 4x − 1)(2 cos 8x − 1) = 1.
Problem 106, May 2009 - Solution
Using the elementary double-angle formula: cos 2θ = 2 cos2 θ − 1, we get
4 cos2 θ − 1 = 2(2 cos2 θ − 1) + 1 = 2 cos 2θ + 1
(∗)
Multiplying both sides of the given equation by 2 cos x + 1 and using (*) 4 times (with
θ = x, 2x, 4x, and 8x, respectively), we get
2 cos x + 1 = (4 cos2 x − 1)(2 cos 2x − 1)(2 cos 4x − 1)(2 cos 8x − 1)
∗
= (2 cos 2x + 1)(2 cos 2x − 1)(2 cos 4x − 1)(2 cos 8x − 1)
= (4 cos 2x − 1)(2 cos 4x − 1)(2 cos 8x − 1)
∗
= (2 cos 4x + 1)(2 cos 4x − 1)(2 cos 8x − 1)
= (4 cos2 4x − 1)(2 cos 8x − 1)
∗
= (2 cos 8x + 1)(2 cos 8x − 1)
∗
= 4 cos2 8x − 1 = 2 cos 16x + 1.
(1)
If 2 cos x + 1 6= 0, then (1) is equivalent to the given equation. Solving (1) we get
2kπ
cos x = cos 16x so 16x = ±x + 2kπ, where k ∈ Z (the set of all integers). Hence, x =
15
2kπ
or
. However, if 2 cos x + 1 = 0, then
17
1
1
1
1
1
cos x = − ⇒ cos 2x = 2 cos2 x − 1 = 2( ) − 1 = − ⇒ cos 4x = − ⇒ cos 8x = − .
2
4
2
2
2
Hence, (2 cos x − 1)(2 cos 2x − 1)(2 cos 4x − 1)(2 cos 8x − 1) = (−2)4 = 16 6= 1. Solving
1
2π
2 cos x + 1 = 0 we find that cos x = − so x = ±
+ 2lπ whose l ∈ Z. Therefore, the
2
3
2kπ
2kπ
2π
solution set of the given equation is {x | x =
or
but x 6= ±
+ 2lπ, k, l ∈ Z}.
15
17
3
Remark: The originally posted solution contained some flaws which were detected by Dr.
Kaiming Zhao. This revised solution is based on his observation and comment. It is easy to
show that the solution set can also be described as:
2kπ 2lπ ,
k, l ∈ Z, k 6≡ 5 or 10 (mod 15)
15 17 Problem 107, June 2009
Prove that if ai ≥ 1, i = 1, 2, · · · , n, then
(1 + a1 )(1 + a2 ) · · · (1 + an ) ≥
2n
(1 + a1 + a2 + · · · + an ).
n+1
Problem 107, June 2009 - Solution
Note first that if xk ≥ 0, k = 1, 2, · · · , n, then
n
Y
(1 + xk ) ≥ 1 +
k=1
n
X
xk .
k=1
Hence,
n
Y
(1 + ak ) = 2
n
k=1
= 2n
n µ
Y
1
k=1
n µ
Y
k=1
Ã
≥ 2n
2
n+1
¶
ak − 1
1+
2
1+
n
=
ak
+
2
2
n
X
ak − 1
à k=1
≥ 2n
1+
n
X
ak − 1
n
X
k=1
!
(ak − 1)
n
=
!
2
k=1
!
n+1
n+1+
Ã
¶
2
n+1
Ã
1+
n
X
k=1
!
ak
.
Problem 108, July 2009
Find all integer solutions of the system of equations xz − 2yt = 3, xt + yz = 1.