Problems Ted Eisenberg, Section Editor

*********************************************************
This section of the Journal offers readers an opportunity to exchange interesting mathematical
problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics,
Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning
proposals and/or solutions can be sent e-mail to <eisenbt@013.net>. Solutions to previously
stated problems can be seen at <http://www.ssma.org/publications>.

————————————————————–

Solutions to the problems stated in this issue should be posted before

March 15, 2021

• 5619: Proposed by Kenneth Korbin, New York, NY

If x, y and z are positive integers such that
x2 + xy + y 2 = z 2
then there are two different Pythagorean triangles with area K = xyz(x + y).
Find the sides of the triangles if z = 61.

• 5620: Proposed by Daniel Sitaru, National Economic College “Theodor Costescu” Drobeta
Turnu-Severin, Mehedinti, Romania
Prove: If a, b, ∈ [0, 1]; a ≤ b, then
 √ab
 
 a √ab r a a+b
s !
 a+b
√ b b
4 ab ≤ a  + +b + ≤ 2(a + b).
a a b b

• 5621: Proposed by Stanley Rabinowitz, Brooklyn, NY

Given non-negative integer n, real numbers a and c with ac 6= 0, and the expression
a + cx2 ≥ 0.
Z

2n+1
Express: a + cx2 2 dx as the sum of elementary functions.

• 5622: Proposed by Albert Natian Los Angeles Valley College, Valley Glen, CA
Suppose f is a real-valued function such that for all real numbers x;
[f (x − 8/15)]2 + [f (x + 47/30)]2 + [f (x + 2/75)]2 =
= f (x − 8/15)f (x + 47/30) + f (x + 47/30)f (x + 2/75) + f (x + 2/75)f (x − 8/15).
     
49 11 1 28 2
If f = , then find f f − f (−42) .
5 3 2 50 25

1
• 5623: Proposed by José Luis Dı́az-Barrero, Barcelona Tech, Barcelona, Spain
Let P be an interior point to an equilateral triangle of altitude one. If x, y, z are the
distances from P to the sides of the triangle, then prove that
x2 + y 2 + z 2 ≥ x3 + y 3 + z 3 + 6xyz.

• 5624: Proposed by Seán M. Stewart, Bomaderry, NSW, Australia

2
tan−1 x − x
Z 1
Evaluate: dx.
0 x2

Solutions

• 5601: Proposed by Kenneth Korbin, New York, NY

Solve: p √
x(x − 1)2 77
= .
(x + 1)2 36

Solution 1 by David A. Huckaby, Angelo State University, San Angelo, TX

Squaring both sides of the equation and cross-multiplying yields 362 x(x−1)2 = 77(x+1)4 .
Expanding gives the quartic equation 77x4 − 988x3 + 3054x2 − 988x + 77 = 0.

Note that the coefficients of the quartic polynomial p on the left side of the equation have
1
a palindrome pattern. So if r 6= 0 is a root of p, then is also a root of p. Indeed, for
  r  
4 1 4 1
x 6= 0 we have x p = p(x). So if r 6= 0 is a root of p, then r p = p(r) = 0, so
  x r
1
that p = 0. (Note that all roots of p are nonzero, since p(0) = 77.)
r

1 988 3 3054 2 988

Let q(x) = p(x) = x4 − x + x − x + 1, and let r1 and r2 be two roots of
77 77 77 77
p and hence also of q. Then
  
1 1
q(x) = (x − r1 )(x − r2 ) x − x−
r1 r2
   
1 1
= (x − r1 ) x − (x − r2 ) x −
r1 r2
   
2 1 2 1
= (x − r1 + x + 1)(x − r2 + x + 1)
r1 r2
     
4 1 1 3 1 1
= x − r1 + + r2 + x + r1 + r2 + + 2 x2
r1 r2 r1 r2
 
1 1
− r1 + + r2 + x+1
r1 r2

2
Equating the x3 (equivalently, the x) coefficients and the x2 coefficients, we obtain the
following system of two equations:

1 1 988
r1 + + r2 + =
r r 77
  1  2
1 1 3054
r1 + r2 + +2=
r1 r2 77

1
Solving for r2 + in the first equation and substituting into the second yields the
r2 
1 2
  
988 1 3054
quadratic equation r1 + − r1 + + − 2 = 0. Since the system
r1 77 r1 77
1 1 1
is symmetric in r1 + and r2 + , the quadratic equation is also true with r1 +
r1 r2 r1
1 58 50
replaced by r2 + . The two solutions of this quadratic equation are and , so that
r2 7 11
1 58 1 50
r1 + = , say, and then r2 + = .
r1 7 r2 11

58 50
So q(x) = (x2 − x + 1)(x2 − x + 1). The solutions to the quadratic equation x2 −
7 √ 11 √
58 29 + 6 22 1 29 − 6 22
x + 1 = 0 are r1 = and = , and the solutions to the quadratic
7 7 r1 √ 7 √
2 50 25 + 6 14 1 25 − 6 14
equation x − x + 1 = 0 are r2 = and = . None of the four is
11 11 r2 11
an extraneous solution to the original equation, so these are its four solutions.

Solution 2 by Bruno Salgueiro Fanego, Viveiro, Spain

p √ p !2 √ !2
x(x − 1)2 77 x(x − 1)2 77 x(x − 1)2 77
2
= ⇒ = ⇒ =
(x + 1) 36 (x + 1)2 36 (x + 1) 4 1296

⇒ 1296x(x2 −2+1) = 77(x4 +4x3 +6x2 +4+1) ⇒ 0 = 77x4 −988x3 +3054x2 3−988x+77 ⇒
⇒ (7x2 − 58x + 7)(11x2 − 50x + 11) = 0 ⇒ 7x2 − 58x + 7 = 0 or 11x2 − 50x + 11 = 0 ⇒

p √
58 ± (−58)2 − 4 · 7 · 7 50 ± 502 − 4 · 11 · 11
⇒x= or x = ⇒
2·7 2 · 11
√ √
29 6 27 25 6 14
⇒x= ± or x = ± .
7 7 11 11
These four numbers are the roots to the given equation.

Solution 3 by Anthony J. Bevelacqua, University of North Dakota, Grand

Forks, ND
We have p √
36 x(x − 1)2 = 77(x + 1)2
and so squaring gives
1296x(x − 1)2 = 77(x + 1)4

3
which yields
1296(x3 − 2x2 + x) = 77(x4 + 4x3 + 6x2 + 4x + 1)
and so finally
77x4 − 988x3 + 3054x2 − 988x + 77 = 0.
Now divide this equation by x2 to find
1 1
77x2 − 988x + 3054 − 988 + 77 2 = 0
x x
and so    
2 1 1
77 x + 2 − 988 x + + 3054 = 0.
x x
1

2
Since x2 + x2
= x + x1 − 2 we have

1 2
   
1
77 x + − 988 x + + 2900 = 0.
x x
Now, by the quadratic formula, we have
1 58 50
x+ = or
x 7 11

which gives
√
2 58 29 ± 6 22
x − x + 1 = 0 with roots x = ≈ 8.16, 0.12
7 7
or √
2 50 25 ± 6 14
x − x + 1 = 0 with roots x = ≈ 4.31, 0.23
11 11
Since each of these roots are positive, our original equation has the four solutions
√ √
29 ± 6 22 25 ± 6 14
,
7 11

Solution 4 by Peter Fulop, Gyomro, Hungary

p √
x(x − 1)2 77
= (1)
(x + 1)2 36

Starting with realign (1) in the following way:

p √
18 4x(x2 − 2x + 1) = 77((x2 − 2x + 1) + 4x) (2)

Let a = 4x and b = x2 − 2x + 1
So we can write that
√ √
18 ab = 77(a + b) (3)

4
 √
r
a
Divided (3) by ab, we get a quadratic equation in
b

√ a a √
r
77 − 18 + 77 = 0 (4)
b b

7

a  11

= (5)
b   11

7
On the other hand
a 4x
= 2 (6)
b x − 2x + 1
Finally from (5) and (6) we have the four roots:
√
22 ± 6 22
x1,2 = 1+
7

√
14 ± 6 14
x3,4 = 1+
11

Solution 5 by Albert Natian, Los Angeles Valley College, Valley Glen, Cali-
fornia.
n √ √ √ √ o
6+√14 6−√14 6+√22 6−√22
Answer. The solution set is 6− 14
, 6+ 14
, 6− 22
, 6+ 22
.

We will first find real solutions and then argue that there can be no other solutions,
not even non-real solutions.

The above equation can be written as

√
x − 1√x
 
77
2 = .
(x + 1) 36
Case One. x > 1.

We have √ √
(x − 1) x 77
= ,
(x + 1)2 36

 √
√ i
   h
x x+1
/ = 77 / [36]
x+1 x−1

which suggests there exists a positive real number m such that

√
x √ x+1
= m 77 and = 36m,
x+1 x−1
5
 36m + 1 72m
x/ (x + 1)2 = 77m2 and x= and x+1= ,
36m − 1 36m − 1
   2
36m + 1 72m
/ = 77m2 ,
36m − 1 36m − 1
(36m − 1) (36m + 1)
= 77m2 ,
(72m)2
362 m2 − 1 = 77 · 722 m4 ,

2
36 36m2 − 1 = 77 · 4 36m2

which, upon the substitution u = 36m2 , becomes

36u − 1 = 308u2 or 308u2 − 36u + 1 = 0 or (14u − 1) (22u − 1) = 0

whose solutions are

1 1
36m2 = u = or 36m2 = u = ,
14 22
1 1
m= √ or m= √
6 14 6 22
36m+1
which, upon insertion into x = 36m−1 , gives
√ √
6 + 14 6 + 22
x= √ or x= √
6 − 14 6 − 22

Case Two. 0 < x ≤ 1.

We have √ √
(1 − x) x 77
= ,
(x + 1)2 36
which, in a manner as in the above,
 √
√ i
   h
x x+1
/ = 77 / [36]
x+1 1−x

which suggests there exists a positive real number m such that

√
x √ x+1
= m 77 and = 36m,
x+1 1−x
36m − 1 72m
x/ (x + 1)2 = 77m2 and x= and x+1= ,
36m + 1 36m + 1
   2
36m − 1 72m
/ = 77m2 ,
36m + 1 36m + 1
(36m − 1) (36m + 1)
= 77m2 ,
(72m)2
362 m2 − 1 = 77 · 722 m4 ,

6

2
36 36m2 − 1 = 77 · 4 36m2

and so, as before, we get

1 1
m= √ or m= √
6 14 6 22

36m−1
which, upon insertion into x = 36m+1 , gives
√ √
6 − 14 6 − 22
x= √ or x= √ .
6 + 14 6 + 22
The solution set of the given radical equation contains at least four different numbers and
is a subset of the solution set of a 4-th degree polynomial equation that can be derived
from the given equation (by squaring both sides of the equation). Since a 4-th degree
polynomial equation has at most 4 (different) solutions, then we can be certain that there
are no other solutions of the given equation and that the solution set of the given equation
is ( √ √ √ √ )
6 + 14 6 − 14 6 + 22 6 − 22
√ , √ , √ , √ .
6 − 14 6 + 14 6 − 22 6 + 22

Comments by other solvers :

David Stone and John Hawkins of Georgia Southern University stated that this problem
is a classic “where does a line intersect a p
hyperbola?” They went √on to say that on their
x(x − 1) 2 77
graphing calculator, the graphs of Y1 = 2
and Y2 = do not even appear
(x + 1) 36
to intersect four times until some significant zooming is done. The curve Y1 starts at the
origin and rises quickly to a maximum of 0.25, which is barely above the horizontal line Y2 ,
then drops quickly to its x-intercept at x = 1. Then it rises again to the same maximum
height of 0.25, barely creeping above Y2 once again, before descending asymptotically
toward the x-axis. (The maximum points of Y1 occur at x = 3 ± 2.)
It is amazing to find such nice solutions. The use of the quadratic formula to find (y, z)
produced rational solutions only because of the numbers chosen in the problem as posed.
How did the poser see all of this?

Comment by Ken Korbin, the proposer:

3696
In problem 5583, the four radii have lengths 16, 49, 9, and 121. And sin A =
4225.
sin A
In problem 5601, if the fraction to the right of the equal sign is replaced by , then,
4
16 49 19 121 sin A 924
the four roots of the equation will be , , , and . N ote : = .
49 16 21 9 4 4225
Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Brian
Bradie, Christopher Newport University, Newport News, VA; Pat Costello,
Eastern Kentucky University, Richmond, KY; Pratik Donga, Junagadh, India;
Eagle Problem Solvers, Georgia Southern University, Statesboro, GA and Sa-
vannah, GA; Farid Huseynov (student; communicated by his instructor Yagub
Aliyev), ADA University, Baku, Azerbaijan; Kee-Wai Lau, Hong Kong, China;
David E. Manes, Oneonta, NY; Ronald Martins, Brazil; Albert Stadler, Her-
rliberg, Switzerland; Seán M. Stewart, Bomaderry, NSW, Australia; David

7
 Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and
the proposer.

• 5602: Proposed by Pedro Henrique Oliveira Pantoja. University of Campina Grande,

Brazil
Prove that:  
 π 3π 
 1 cos sin 
 7 7 
 √
 

3π 2π π  = 7.
 
det sin sin sin2
 7 7 7

 8
 
 π π 
 0 tan 2 sin2 
7 7
 

Solution 1 by Albert Stadler, Herrliberg, Switzerland

π π √ 2π 3π π x
Let x = sin , y = cos = 1 − x2 . Then sin = 2xy, sin = 3x − 4x3 , tan = .
7 7 7 7 7 y
Let d be the value of the determinate. We expand the determinant along the first column
and get
2π π  π 3π 
   
sin sin2   cos sin
 

 7 7  3π  7 7 
d = det   − sin det  =
π π
 7 π π

 tan 2 2
2 sin  tan 2 sin 
   
7 7 7 7
 
2 π 2π π 2 2π 3π 2 π π π 3π
= 2 sin sin − tan sin − sin 2 sin cos − tan sin =
7 7 7 7 7 7 7 7 7

x3 3x2 − 4x4 2x3 (4 − 12x2 + 8x4 − y 2 + 4x2 y 2 )

 
= 4x3 y − − (3x − 4x3 ) 2x2 y − = =
y y y

2x3 (4 − 12x2 + 8x4 − (1 − x2 ) + 4x2 (1 − x2 )) 2(x − 1)x3 (x + 1)(4x2 − 3)

= = =
y y

r r
2y 2 x2 (3x − 4x3 ) π 2π 3π π 2π 3π 4π 5π 6π 7
= = sin sin sin = sin sin sin sin sin sin =
y 7 7 7 7 7 7 7 7 7 64

since for all integers n ≥ 2

n−1   n−1 Pn−1 πik n−1

Y kπ Y  πik −πik
 Y 2πik

2 sin = (−i) e n − e n = (−i)n−1 e k=1 n 1 − e− n =
n
k=1 k=1 k=1

n−1
πi(n−1) Y 2πik
 xn − 1
= (−i)n−1 e 2 1−e n = lim = n, (see problem 5497, April 2018).
x→1 x − 1
k=1

8
Solution 2 by Brian Bradie, Christopher Newport University, Newport News,VA
Subtracting sin 3π
7 times the first row from the second row yields

cos π7 sin 3π cos π7 sin 3π

   
 1   1 
7 7
det  sin 3π sin 2π sin2 π7  = det  0 sin − cos π sin 3π
2π 2 π 3π
   
7 7   7 7 7 sin 7 − sin2 7
.

 0 tan π7 2 sin2 π7   0 tan π7 2 sin2 π7 

Now,
 
2π π 3π 3π π π 3π
sin − cos sin = sin − − cos sin
7 7 7 7 7 7 7
3π π 3π π π 3π 3π π
= sin cos − cos sin − cos sin = − cos sin
7 7 7 7 7 7 7 7
and
  
π
2 3π π 3π π 3π
sin − sin2 = sin − sin sin + sin
7 7 7 7 7 7
  
π 2π 2π π 2π 4π
= −2 sin cos 2 sin cos = − sin sin ,
7 7 7 7 7 7
so
cos π7 sin 3π
 
 1 
7
det  sin 3π sin 2π sin2 π7  = −2 sin3 π cos 3π + tan π sin 2π sin 4π
 
7 7  7 7 7 7 7
 0 tan π7 2 sin2 π7 
sin π7
 
2 π π 3π 2π 4π
= −2 sin cos cos + sin sin
cos π7 7 7 7 7 7
sin π7 sin 2π
 
7 π 3π 4π
= π − sin cos + sin
cos 7 7 7 7
π 2π  
sin 7 sin 7 π 3π 3π π 3π π
= − sin cos + sin cos + cos sin
cos π7 7 7 7 7 7 7
π 2π 3π
= sin sin sin .
7 7 7
To show √
π 2π 3π 7
sin sin sin = ,
7 7 7 8
let n ≥ 2 be an integer. The roots of z n − 1 are ωk = e2ikπ/n , for k = 0, 1, 2, . . . , n − 1.
Then
n−1
X n−1
Y
z n − 1 = (z − 1) z k = (z − 1) (z − ωk ),
k=0 k=1
or
n−1
X n−1
Y
zk = (z − ωk ).
k=0 k=1

Substituting z = 1 yields
n−1
Y
n= (1 − ωk ).
k=1

9
Next,
  r
 2kπ 2kπ  = 2 − 2 cos 2kπ = 2 sin kπ ,

|1 − ωk | = 1 − cos − i sin
n n  n n
so
n−1 n−1
Y
n−1
Y kπ
n = |n| = |1 − ωk | = 2 sin .
n
k=1 k=1
Take n = 7 and note
4π 3π 5π 2π 6π π
sin = sin , sin = sin , and sin = sin .
7 7 7 7 7 7
It follows that
√
6π 2 2π 3π π 2π 3π 7
7 = 2 sin sin2 sin2 or sin sin sin = .
7 7 7 7 7 7 8

Editor0 s comment : The solution submitted by Seán M. Stewart of Bomaderry,

Australia started off by proving three identities:
π 2π 3π 1
1. C = cos − cos + cos =
7 7 7 2√
π 2π 3π 7
2. S = − sin + sin + sin =
7 7 7 2
sin π7 + 2 sin 3π
7
√
3. T = = 7.
cos π7

The proof of C was straight forward, in proving S he first showed that the square of the
LHS equals the square of the RHS,√and he then chose the positive square roots; in proving
7 √
S
T he showed that T = = 2 1 = 7.
1−C 1− 2
T
He then expanded the Determinate D down the third column and showed that D = =
√ 8
78. Lots of algebra, but it worked.

Solution 3 by Kee-Wai Lau, Hong Kong, China

π
Let α = . The given determinate, denoted by D equals
7
2 sin 2α sin2 α + sin2 3αα − tan α sin2 α − 2 sin2 α cos α sin 3α
4 sin3 α cos2 α + sin2 3α sin α − sin3 α − 2 sin2 α cos2 α sin 3α
= .
cos α

Let k = sin α. By using the relations cos2 α = 1 − k 2 and sin 3α = 3k − 4k 2 ,

we see that the numerator of D equals

8k 7 − 14k 5 + 6k 3 = 2k 3 (1 + k)(1 − k)(3 − 4k 2 ).

√
π 1 7
Since 0 < k < sin = , so D > 0. Hence to prove that D = , it suffices to show that
6 2 8
7
D2 = or 64(8k 7 − 14k 5 + 6k 3 )2 − 7(1 − k 2 ) = 0, or
64
10
 (k + 1)(k − 1)(64k 5 − 48k 4 − 8k 2 − 1)(64k 6 − 112k 6 + 56k 2 − 7) = 0. (1)

It is well known that sin 7θ = − sin θ(64 sin6 θ − 112 sin4 θ + 56 sin2 θ − 7)
− sin 7α
for any real number θ. Hence, 64k 6 − 112k 4 + 56k 2 − 7 = = 0.
k
Thus (1) holds and this completes the solution.

Editor0 s comment : David Stone and John Hawkins of Georgia Southern Uni-
versity used a statement  in  their
√ solution that was proved by P dilip k Stefan V.,
π  3π 7
sin 2π


that sin 7 sin = (see: (https://socratic.org/questions/how-do-you-
7   7  8
π 2π 3π
evaluate sin sin sin ). They concluded their solution by stating “in this
7 7 7    
2π 3π
problem, the surprise is that the given determinate equals sin π7 sin sin ).”
7 7
Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Bruno
Salgueiro Fanego, Viveiro, Spain; Peter Fulop, Gyomro, Hungary; Seán M.
Stewart, Bomaderry, NSW, Australia; David Stone and John Hawkins, Geor-
gia Southern University, Statesboro, GA, and the proposer.

• 5603: Proposed by Michael Brozinsky, Central Islip, NY

In an election 50 votes were cast for candidate A and 50 for candidate B. The candidates
decide to end the tie as follows; by tallying the votes at random and if A is ever in the
lead by 3 votes, then Candidate A will be declared the winner. Otherwise Candidate B
wins. What is the probability that A wins?

Solution 1 by Albert Stadler, Herriliberg, Switzerland

Consider the 2-dimensional integer lattice Z 2 in the Euclidean space R2 whose lattice
points are 2-tuples of integers. Every random tallying of votes can be mapped bijectively
to a path from (0, 0) to (50, 50) under the rule that if starting at (0, 0) and if a vote for
A is picked a step from (x, y) to (x + 1, y) is done  and if a vote for B is picked a step
100
from (x, y) to (x, y + 1) is done. In total there are paths from (0, 0) to (50, 50).
50
By the reflection principle, the number of paths from (0, 0) to (50, 50) where A will be in
the lead by 3 votesat some point is equal to the number of paths from (0, 0) to (47, 53)
100
which equals . Therefore the probability that A wins equals
47
 
100
47 100!50!50! 48 · 49 · 50 9800
 = = = ≈ 83.7%.
100 47!53!100! 51 · 52 · 53 11713
50

Solution 2 by Albert Natian, Los Angeles College, Valley Glen, CA

48·49·50
Answer: 51·52·53 ≈ 0.836677.

11
First we generalize the problem as follows:

In an election m votes are cast for candidate A and n for candidate B. The candidates
agree to break the tie as follows: the votes will be tallied uniformly at random and if A
is ever in the lead (for the first time) by L votes, then A is declared the winner; oth-
erwise B wins. In this generalization, unlike the original statement of the problem, an
additional advantage/disadvantage of i votes is initially accorded candidate A. That is,
if, say, i = 1 and L = 3, and the first two tallies are for A, then A wins. But if the first
tally is for B, then A’s 1-point advantage is lost (and so i becomes 0) and now A needs
to lead B by 3 in order to win, and if the following tally is again for B, then now A’s
advantage/disadvantage i becomes −1. So i measures A’s lead over B by the latest tally.
It’s clear that i can be negative, zero or positive. So when (if ever) i becomes L, then A
wins. What is the probability that A wins?

We let P (m, n, i, L) denote the probability that A wins. It’s immediate that P satis-
fies the following conditions

P (m, n, i, L) = 1 if L = i or i=n+L−m

P (m, n, i, L) = 0 if m < L − i
L − i + n −1
 
P (L − i, n, i, L) = .
n
The first tally is a vote for A with probability m/ (m + n) and a vote for B with probability
n/ (m + n). If the first tally is a vote for A, then there will now be (m − 1) votes remaining
for A and n votes for B. Also if the first tally is a vote for A, then i is incremented by 1.
However, if the first tally is a vote for B, then there will now be (n − 1) votes remaining
for B and m votes for A. Also if the first tally is a vote for B, then i is decremented by 1.
Se we assert
m n
P (m, n, i, L) = · P (m − 1, n, i + 1, L) + · P (m, n − 1, i − 1, L) .
m+n m+n
In order to solve the above recursion with the given conditions, we define f as
 
m+n
f (m, n, i, L) = P (m, n, i, L) .
m

So
m + n −1
 
P (m, n, i, L) = f (m, n, i, L) .
m

12
Now
   
m+n m+n
f (m, n, i, L) = P (m, n, i, L) = P (m, n, i, L)
m n

  
m+n m n
= · P (m − 1, n, i + 1, L) + · P (m, n − 1, i − 1, L)
m m+n m+n

   
m+n m m+n n
= · P (m − 1, n, i + 1, L) + · P (m, n − 1, i − 1, L)
m m+n n m+n

   
m−1+n m+n−1
= · P (m − 1, n, i + 1, L) + · P (m, n − 1, i − 1, L)
m−1 n−1

= f (m − 1, n, i + 1, L) + f (m, n − 1, i − 1, L) .

The conditions for f are

 
m+n
f (m, n, i, L) = if L = i or i = n + L − m
m

f (m, n, i, L) = 0 if m < L − i
P (L − i, n, i, L) = 1.
The Solution for the recursion

f (m, n, i, L) = f (m − 1, n, i + 1, L) + f (m, n − 1, i − 1, L)

satisfying the aforementioned conditions is

   
m+n m+n
f (m, n, i, L) = = .
n+L−i m−L+i

Thus
 −1  
m+n m+n m! n!
P (m, n, i, L) = = .
m m−L+i (m − L + i)! (n + L − i)!

To answer the original question, we let m = 50, n = 50, i = 0, L = 3. Then

50! 50! 48 · 49 · 50
P (50, 50, 0, 3) = = ≈ 0.836677.
47! 53! 51 · 52 · 53

Solution 3 by Moti Levy, Rehovot, Israel

The process of counting 2N votes can be modeled as a random walk from (0, 0) to (2N, 0)
with up-steps and down-steps of one unit each. The paths are called grand-Dyck path
(or free-Dyck path).
The number of all grand-Dyck paths from (0, 0) to (2N, 0) is equal to 2N


N . Clearly all
paths have equal probability.

13
Let us denote by Nn the number of “losing paths”, i.e., grand-Dyck paths from (0, 0) to
(n, 0) that never exceed the height 2.
Then the probability that candidate A will be declared the winner is
N2N
P (A wins) = 1 − 2N

.
N

Let Ψ (z) be the generating function of the sequence of numbers of grand-Dyck paths from
(0, 0) to (n, 0) that never exceed the height 2.
∞
X
Ψ (z) = Nn z n
n=0

The article by Panny and Prodinger [1] , states that (see theorem 4.1, page 130) the
generating function is

1 + v2 v
1 − v6 ,


Ψ (z) = 2
z= ,
1−v 1 + v2
or
1  p 
Ψ (z) = 1 + 2v 2 + 2v 4 + v 6 , 1 − 1 − 4z 2 .
v=
2z
Now we find several generating functions on our way to evaluate Ψ (z) .
1
1
(− 1 )(− 32 )···(− 2n−3
2 )
By definition of the binomial coefficient n2 = 2 2 n! , hence

1   1 n = 0,
1
2 = 2 n = 1, (1)
n  (−1)n−1 1 (2n−2
 n )
n−1 22n−1 n > 1.
By the binomial theorem and (1),
∞ 1
2

1 X m 2m
2 z 2m ,
1 − 4z =1+
2
(−1) 2 (2)
m
m=1
∞ 3

3
2 2
X m 2m 2
1 − 4z =1+ (−1) 2 z 2m
m
m=1
∞
X 1 ∞
X 1
2 m 2m 2 2m 2 m 2m 2
= 1 − 4z + (−1) 2 z − 4z (−1) 2 z 2m (3)
m m
m=1 m=1

1  2
3
p 
1+2v 2 +2v 4 +v 6 = − 8z 1 − 4z 2 2
+ 40z 4
− 4z 2
+ 3 1 − 4z 2 + 32z 2 − 24z 4 − 8
16z 6
(4)
Substituting (2) and (3) in (4) and after tedious simplifications we get,
∞
X  1   1   1 
m 2m+1
Ψ (z) = 1 + (−1) 2 2 +8 2 + 16 2 z 2m
m+1 m+2 m+3
m=1
∞ 
9m2 + 9m + 6

X 2m
= z 2m .
m m3 + 6m2 + 11m + 6
m=0

14
It follows that
9N 2 + 9N + 6
 
2N
N2N = , (5)
N N 3 + 6N 2 + 11N + 6
and that
9N 2 + 9N + 6
P (A wins) = 1 − .
N 3 + 6N 2 + 11N + 6
9∗502 +9∗50+6 ∼
For the case N = 50, we have P(A wins) = 1 − 503 +6∗502 +11∗50+6 = 0.836 68.
Remarks:
1) Equation (5) can be verified for small values of N by manual counting of the paths:
2N


N N2N N P (A wins)
1 2 2 0
2 6 6 0
1
3 19 20 20 = 0.05
8 ∼
4 62 70 70 = 0.114 29

2) For large N
  
9 1
P (A wins) = 1 − 1+O
N N

3) We give here the essential steps in Panny and Prodinger derivation (see [2]).
We consider grand-Dyck paths form (0, 0) to (n, 0). We allow the path to touch −h and
k but not −h − 1 and k + 1. Let Nn,h be the number of paths which do not touch −h − 1
and k + 1 and lead to level i. Let Ψi be the generating function of the sequence (Nn,h )∞
n=0 ,

∞
X
Ψi = Nn,h z n .
n=0

Looking at the last step of the paths we write the following recurrences:

Nn,−h = Nn−1,−h
Nn,−h+1 = Nn−1,−h + Nn−1,−h+2
..
.
Nn,−1 = Nn−1,−2 + Nn−1,0

Nn−1,−1 + Nn−1,1 if n > 0
Nn,0 =
1 if n = 0
Nn,1 = Nn−1,0 + Nn−1,2
..
.
Nn,k−1 = Nn−1,k−2 + Nn−1,k
Nn,k = Nn−1,k

15
which implies (in terms of the generating functions)
   
Ψ−h 0
1 −z 0 · · · · · · 0
 
 ..   .. 
 −z 1 −z 0 · · · 0  .   . 
  ..
 
..

 0 −z 1 −z · · · 0  .
 
.

 = (6)
   
··· ···  
Ψ0 1

    
 0 · · · 0 −z 1 −z 
 ..
 
  ..


0 · · · · · · 0 −z 1
 .   . 
Ψk 0

Of course, we are interested in Ψ0 .

The matrix is tridiagonal, hence its determinant fn of n × n matrix can be found by
solving the recurrence relation (see [3])

fn = fn−1 − z 2 fn−2 .
 2  2
v 1 v2
Setting z = (1+v)2
, for convenience, the solution is fn = a v 2 +1
+b v 2 +1
. Applying
the initial conditions f1 = 1 and f2 = 1 − z 2 , we find that
1 1 − v 2n+2
fn = .
1 − v 2 (v 2 + 1)n
Applying Cramer’s rule, to solve (6), we get

fh fk 1 + v2  2k+2
 v 2h+2 − 1
Ψ0 = = 1 − v
fh+k+1 1 − v2 v 2h+2k+4 − 1
Now we send h to -∞, to get
1 + v2  2k+2

Ψ0 (z) = 1 − v .
1 − v2
References:
[1] Wolfgang Panny, Helmut Prodinger, “The expected height of paths for several notions
of height”, Studia Scientiarum Mathematicarum Hungarica 20 (1985), 119-132.
[2] Helmut Prodinger, “The number of restricted lattice paths revisited ”, Filomat 26:6,
1133-1134, published by Faculty of Sciences and Mathematics, University of Niŝ, Serbia.
[3] Wikipedia, ”Tridiagonal Matrix ” entry.

Solution 4 by Kee-Wai Lau, Hong Kong, China

9800
We show that A wins with probability .
11713
Suppose, in general, that A obtained a votes and B obtained b votes with a ≥ b.
Denote by P0 (a, b) = the probability that there will ever be a tie in a full tallying. For
k = 1, 2, 3, let Pk (a, b) be the probability that B will ever lead by k votes in a full
2b
tallying. It is known (1), pp. 6, 37, 38, Problem 22: The ballot box that P0 (a, b) = .
a+b
We use this result to find P1 (a, b), P2 (a, b), and P3 (a, b)
a
successively. The probability that A gets the first vote equals and the
a+b
b
probability that B gets the first vote equals . Hence by conditioning on the first
a+b
16
 2b a b b
vote, we obtain = P0 (a, b) = P1 (a − 1, b) + / . Thus P1 (a − 1, b) =
a+b a+b a+b a
b
and P1 (a, b) = . By conditioning again, we have
a+1
b b b(b − 1)
= P1 (a, b)P2 (a − 1, b) + , giving P2 (a − 1, b) = and
a+1 a+b a(a + 1)
b(b − 1)
P2 (a, b) = . Finally, we have
(a + 1)(a + 2)
b(b − 1) a
= P2 (a, b) = P3 (a − 1, b) +
(a + 1)(a + 2) a+b
b(b − 1)(b − 2) b(b − 1)(b − 2)
This gives P3 (a − 1, b) = and P3 (a, b) ,
a(a + 1)(a2 ) (a + 1)(a + 2)(a + 3)
If a = b, then P (A is ever in the lead by 3 votes)= P (B is ever in the lead by 3 votes)

= P3 (a, a),

a(a − 1)(a − 2)
= .
(a + 1)(a + 2)(a + 3)

By putting a = 50, we obtain the result stated at the beginning.

Reference: 1. F. Mosteller: Fifty Challenging Problems in Probability with Solutions,
Dover Publications, Inc., 1987.

Solution 5 by the Eagle Problem Solvers, Georgia Southern University, States-

boro, GA and Savannah, GA

Let S be the set of all sequences of 50A0 s and 50B 0 s, and let A be the subset of S
consisting of the sequences in which there is an initial subsequence with three more A0 s
than B 0 s. We claim that there is a one-to-one correspondence between A and the set B
of all sequences of 47A0 s and 53 B 0 s.
Let (x1 , x2 , · · · , x100 ) be a sequence in A, and let k be the smallest integer for which the
initial sequence (x1 , x2 · · · , xk ) has exactly three more A0 s than B 0 s. Then the initial
sequence (x1 , x2 , · · · , xk ) has b B 0 s and b + 3 A0 s for some integer b, where 0 ≤ b ≤ 47,
and the remaining sequence (xk+1 , xk+2 , · · · x100 ) will have 50 − bB 0 s and 47 − bA0 s. Now
transform the sequence (x1 , x2 , · · · , x100 ) by changing the A0 s to B 0 s and the B 0 s to A0 s
in the initial k terms, and leaving the remaining 100 − k terms unchanged. This new
sequence will have b + (47 − b) = 47A0 s and (b + 3) + (50 − b) = 53B 0 s, and thus will be
a sequence in B.
In the other direction, if we begin with a sequence (y1 , y2 , · · · , y100 ) of 47A0 s and 53B 0 s,
then let k be the smallest integer for which the initial sequence (y1 , y2 , · · · yk ) has exactly
three more B 0 s than A0 s. Change the A0 s to B 0 s and B 0 s to A0 s in this initial sequence
and leave the tail end of the sequence unchanged. The resulting sequence will have 50A0 s
and 50B 0 s, and the initial sequence of k terms will have exactly three more A0 s than B 0 s.
 
100
Thus, the sets A and B have the same cardinality, |A| = |B| = and the probability
47

17
 that Candidate A will win is
 
100
|A| 47
=  
|B| 100
50

100! 50!50!
= ·
47!53! 100!
50 · 49 · 48
=
51 · 52 · 53
9800
=
11, 713
≈ 0.836677.

More generally, suppose A is declared the winner if A is ever in the lead by k votes,
where k is a positive integer less than or equal to 50. Let A be the subset of S containing
sequences in which there is an initial subsequence containing exactly k more A0 s than B 0 .
As before, there is a one−to−one correspondence between  A and the  set B of sequences
0 0 100
with 50 − kA s and 50 + kB s. Thus, the cardinality of A is and the probability
50 − k
that A is declared the winner is
 
100
50 − k (50!)2 50 · 49 · · · (51 − k)
  = = .
100 (50 − k)!(50 + k)! 51 · 52 · · · (50 + k)
50

1
The value of k for which this probability comes closest to is k = 6, for which the
2
probability is

189, 175
≈ 0.48942.
386, 529

Also solved by the proposer.

• 5604: Proposed by Albert Natian, Los Angeles Valley College, Valley Glen, CA
Prove:
  n−1
N 1 X −irµ 2π  2π N

= lim e n 1 + eiµ n
r n→∞ n
µ=0

where N, r ∈ N and i2 = −1.

Solution 1 by Albert Stadler, Herrliberg, Switzerland

The sum on the right hand side is a Riemann sum that converges to
Z 1 N  Z 1  
−2πirx 2πix N
X N −2πirx+2πirnx N
e (1 + e ) dx = e dx = ,
0 n 0 r
n=0

18
as claimed. We have used the binomial theorem and the fact that for an integer k
Z 1 
2πikx 1, k = 0
e dx =
0 0, k 6= 0.

Solution 2 by Brian Bradie, Christopher Newport University, Newport News,VA

2π
Let θ = n . By the binomial theorem,

N  

iµθ
N X N iµkθ
1+e = e ,
k
k=0

so
n−1 n−1 N   N   n−1
X
−irµθ

iµθ
N XX N iµθ(k−r) X N X iµθ(k−r)
e 1+e = e = e .
k k
µ=0 µ=0 k=0 k=0 µ=0

Now, for r > N ,  

N
= 0,
r
and, for each n > r,
n−1
X 1 − einθ(k−r) 1 − e2πi(k−r)
eiµθ(k−r) = = = 0,
µ=0
1 − eiθ(k−r) 1 − eiθ(k−r)

so
n−1  
1 X −irµ 2π  iµ 2π N
 N
lim e n 1+e n =0= .
n→∞ n r
µ=0

Finally, for r ≤ N and for each n > N

N   n−1   n−1 N   n−1
X N X N X X N X iµθ(k−r)
eiµθ(k−r) = 1+ e
k r k
k=0 µ=0 µ=0 k=0,k6=r µ=0
N
N 1 − einθ(k−r)
   
N X
= n +
r k 1 − eiθ(k−r)
k=0,k6=r
 
N
= n ,
r
so
n−1  
1 X −irµ 2π  2π N
 N
lim e n 1 + eiµ n = .
n→∞ n r
µ=0

Solution 3 by Seán M. Stewart, Bomaderry, NSW, Australia

Let
n−1  N
X 2π 2π
S(r, N ) = e−irµ n 1 + eiµ n ,
µ=0

19
where r, N ∈ N . To prove the result given we also need to make the additional assumption
that r ≤ N . From the binomial theorem we can write the term appearing in the brackets
inside the sum as
N  

iµ 2π
N X N 2πiµk
1+e n = e n .
k
k=0
Thus
N   n−1
X N X
S(r, N ) = eµθk ,
k
k=0 µ=0
2πi
after the order of the summations have been interchanged. Here θk = n (k − r). As
n−1
X enθk − 1
eµθk = ,
eθk − 1
µ=0

we see that
n−1
X e2πi(k−r) − 1
fk = eµθk = 2πi = 0,
(k−r)
µ=0 e n −1

provided k 6= r since e2πi(k−r) = 1 as k is a non-negative integer and r a positive integer.

When k = rN we have
n−1
X n−1
X
fk = eµθk = 1 = n.
µ=0 µ=0

So for r ≤ N where r, N ∈ N we have

(
0, k 6= r,
fk =
n, k = r.

Thus
N    
X N N
S(r, N ) = fk = n .
k r
k=0
So for the desired limit we have
n−1    
1 X −irµ 2π  iµ 2π N
 1 N N
lim e n 1+e n = lim ·n = ,
n→∞ n n→∞ n r r
µ=0

as required to prove.

Solution 4 by Ulrich Abel, Technische Hochschule Mittelhessen, Germany

Let N, r ∈ N . Applying the binomial formula we obtain
n−1 N   n−1
X
−irµ 2π

iµ 2π
N X N X 2π
e n 1+e n = eiµ n (k−r) .
k
µ=0 k=0 µ=0

If k = r, the inner sum has value equal to 1. Otherwise, the formula for the geometric
sum tells us that its value is equal to

1 − e2πi(k−r)
= 0,
1 − e2πi(k−r)/n
20
provided that n > |k − r|. Hence,
n−1  
1 X −irµ 2π  iµ 2π N
 N
lim e n 1+e n = .
n→∞ n r
µ=0

Solution 5 by Moti Levy, Rehovot, Israel

n−1  2π
 Z 1
1 X −irµ 2π 1+eiµ n
N
lim e n
= e−ir2πx 1 + ei2πx dx
n→∞ n 0
µ=0
Z 1 N  
X N
= e−ir2πx ei2πkx dx
0 k
k=0
N  Z 1
X N
= e−i2πrx ei2πkx dx
k 0
k=0
Z 1 
1 if k=r
e−i2πrx ei2πkx dx =
0 0 if k 6= r
and the result follows.

Solution 6 by Peter Fulop, Gyomro, Hungary

n−1  N
1 P −irµ 2π 2π N
+ eiµ n


Prove S = lim e n 1 =
n→∞ n µ=0 r

where N, r ∈ N
Riemann sum
We know that Riemann sum gives us the following formula for a function f ∈ C 1 :

n Z1
1X µ
lim f( ) = f (x)dx (8)
n→∞ n n
µ=0 0

In our case:

n Z1
1 X −irµ 2π  2π N
 2N  N
lim e n 1 + eiµ n − lim = e−2πixr 1 + e2πix dx (9)
n→∞ n n→∞ n
µ=0 | {z } 0
→0

Contour integral
Applying the z = e2πix substitution in the integral of (2) we get the following contour
integral:

(1 + z)N
I
1
S= dx (10)
2πi z r+1
|z|=1

21
 N
Using the Cauchy’s differentiation formula and realized that g(z) = (1+z)
z r+1
function has
r + 1 poles at z = 0 which are inside the unit circle |z| = 1, so we get:

(1 + z)N 1 dr
I
1
S= dx = (1 + z)N  (11)
2πi z r+1 r! dz r 
|z|=1 z=0

Performing the derivates ((1 + z)N can be differentiated r times):

1 dr 1 
S= (1 + z)N  = N (N − 1)(N − 2)....(N − r + 1) (1 + z)N −r  (12)
r! dz r  r! 
z=0 z=0

N!
Easy to realize that N (N − 1)(N − 2)....(N − r + 1) = (N −r)!
Finally substitute back to (5) we give the result:

(1 + z)N
I  
1 N
S= dx = (13)
2πi z r+1 r
|z|=1

So the statement is proved.

Also solved by Pratik Donga, Junagadh, India; Kee-Wai Lau, Hong Kong,
China, and the proposer.

• 5605: Proposed by José Luis Dı́az-Barrero, Barcelona Tech, Barcelona, Spain

Let b and c be distinct coprime numbers. Find the smallest positive integer a for which

gcd(ab − 1, ac − 1) = 100.

Solution 1 by Albert Stadler, Herrliberg, Switzerland

Let n, t, s be natural numbers. We claim that

gcd(nr − 1, ns − 1) = ngcd(r,s) − 1. (*)

For the proof we proceed by induction on max(r, s). The statement is trivial for max(r, s) =
1 and r = s. Suppose the statement holds true for all r, s with max(r, s) ≤ m. Suppose
that r = m + 1 > s. Then

gcd(nr − 1, ns − 1) = gcd(nr − 1 − (ns − 1), ns − 1) = gcd(ns (nr−s − 1), ns − 1) =

= gcd(nr−s − 1, ns − 1) = ngcd(r−s,s) − 1 = ngcd(r,s) − 1,

which concludes the proof (∗).

22
Clearly, b ≥ 0, c ≥ 0. By (∗) 100=gcd(ab − 1, ac − 1) = agcd (b,c) − 1 = a − 1. Hence
a = 101.

Solution 2 by Pratik Donga, Junagadh, India

We can use the following formula to find a: gcd nb − 1, nc − 1 = ngcd(b,c) − 1. Here we

let n = a and so we obtain gcd ab − 1, ac − 1 = agcd (b,c)−1=100 . But gcd (b, c) = 1 since

b and c are coprime we have a − 1 = 100 → a = 101.

Editor0 s Comment : Kee-Wai Lau of Hong Kong, China noted that the proof of this
formula can be found in T. Andreescu, D. Andrica, and Feng, Z. 104 Number Theory
Problems, From the Training of the USA IMO Team, Birkhauser, 2007, on page 112 as
part of the solution to problem 38 on page 79.

Solution 3 by David E. Manes, Oneonta, NY

More generally, if a, b and c are positive integers,then

gcd(ab − 1, ac − 1) = agcd(b,c) − 1 (1)

Therefore, if gcd(b, c) = 1, (b 6= c), then

gcd(ab − 1, ac − 1) = agcd(b,c) − 1 = a − 1 = 100

so that the only positive integer satisfying the given equation is a = 101.
To prove equation (1), we will us the following result: Let m be a positive integer and a
and b are integers relatively prime to m. If x and y are integers such that ax ≡ bx (mod
m) and ay ≡ by (mod m), then agcd(x,y) ≡ bgcd(x,y) (mod m)
Since gcd(b, c) divides both b and c, the polynomial xgcd(b,c) − 1 divides both xb − 1
and xc − 1. Hence, agcd(b,c)−1 divides both ab − 1 and ac − 1 so that agcd(b,c) − 1 is a
divisor of gcd(ab − 1, ac − 1). On the other hand, if m divides both ab − 1 and ac − 1,
then gcd(a, m) = 1 and ab ≡ 1 ≡ 1b (mod m). By the above stated result, it follows
that agcd(b,c) ≡ 1 (mod m); that is to say, m is a divisor of agcd(b,c) − 1. Therefore,
gcd(ab − 1, ac − 1) is a divisor of agcd(b,c) − 1. Hence, agcd(b,c) − 1 = gcd(ab − 1, ac − 1).

Solution 4 by David Stone and John Hawkins, Georgia Southern University,

Statesboro, GA
We shall show that a = 101.
Note that an − 1 = (a − 1) an−1 + an−2 + . . . + a + 1 .

am+1 − 1
For convenience, let Pn = = am + am−1 + . . . + a + 1. Note that Pn has m + 1
a−1
terms.
Thus gcd ab − 1, ac − 1 = gcd ((a − 1)Pb−1 ) , (a − 1)Pc−1 = (a − 1) gcd(Pb−1 , Pc−1 ).

In Comment 1 below, we will use the Euclidean Algorithm to show that, for coprime b
and c, we have gcd(Pb−1 , Pc−1 ) = 1. Hence we have gcd ab − 1, ac − 1 = a − 1.

That is, 100 = gcd ab , ac = (a − 1) · 1 = a − 1, so a = 101.

Comment 1: The Euclidean Algorithm process which is used to determine that gcd (b, c) =
1 can be used as a guide to show that gcd (Pb−1 , Pc−1 ) = 1.

23
 We demonstrate with an illustrative example.
Let b = 25 and c = 9.
Apply the Euclidean Algorithm:

25 = 2 · 9 + 7
(1) 9=1·7+2
7=3·2+1

The final remainder is 1, verifying that gcd(25, 9) = 1.
Now we want to demonstrate that gcd(P24 , P8 ) = 1 by applying the Euclidean Algorithm.
The first step in (1) above tells us exactly what to do as a first step here. Divide P24 by
P8 : P8 goes 2 times with a remainder of 7. That is, the 25 terms of P24 a are grouped into
2 blocks of (9 terms each ) and there are 7 terms left over.
P24 = a24 + a23 + a22 + . . . + a + 1 =

a24 + a23 + a22 + a21 + a20 + a19 + a18 + a17 + a16 +

=

a15 + a14 + a13 + a12 + a11 + a10 + a9 + a8 + a7 +

+

+ a6 + a5 + a4 + a3 + a2 + a1 + 1 =

= a16 a8 + a7 + a6 + a5 + a4 + a3 + a2 + a + 1 +

+ a7 a8 + a7 + a6 + a5 + a4 + a3 + a2 + a + 1 +

+ a6 + a5 + a4 + a3 + a2 + a + 1 =

a16 + a7 P8 + P6 .


=

For the next step, we divide P8 by P6 . The second step of (1) above tells us exactly what
happens: P6 goes 1 times with a remainder 2. That is the 9 terms of P8 are grouped into
1 block (of 7 terms) and there are 2 terms left over.

P8 = a8 + a7 + a6 + a5 + a4 + a3 + a2 + a + 1 =

a8 + a7 + a6 + a5 + a4 + a3 + a2 + (a + 1) =


=

= a2 +a6 + a5 + a4 + a3 + a2 + a + 1 + (a + 1) =

= a2 P6 + P1 .

For the third step, we divide P6 by P1 . The third step of (1) above tells exactly what happens:
P1 goes 3 times with a remainder 1. That is, the 7 terms of P6 are grouped into 3 blocks of 2
terms, and there is 1 term remaining (with must be a 1):
P6 = a6 + a5 + a4 + a3 + a2 + a + 1

24
 a6 + a5 + a4 + a3 + a2 + a) + 1



=

= a5 (a + 1) + a3 (a + 1) + a (a + 1) + 1

a5 + a3 + a (a + 1) + 1


=

a5 + a3 + a P1 + 1.


=

Summarizing this Euclidean Algorithm:

a15 + a7 P8 + P6


P24 =

P8 = a2 P6 + P1

P6 = a5 + a3 + a)P1 + 1. Therefore,

gcd (P24 , P8 ) = 1.

This process can be formalized. For any coprime b and c, the Euclidean Algorithm can be
employed to show that gcd (Pb−1 , Pc−1 ) = 1; the process demonstrated above will work and
terminate with 1, by repeatedly applying the following Lemma.
Lemma: For 1 ≤ n < m, when Pm is divided by Pn , the remainder is a Pk , with 0 ≤ k < n.
Proof: Use the Division Algorithm to divide m + 1 by n + 1:
m + 1 = q(n + 1) + r where 0 ≤ r ≤ n + 1. Then it is straight forward to see that
Pm = Q · Pn + Pr−1 ,where Q = qj=1 xm+1)−j(n+1) .
P

Note that the operations with the Pn can be considered as with integers or as with polynomials
with integer coefficients.

Comment 2: It seems surprising that the answer is independent of the values of b and c. What
if b and c are not relatively prime? Suppose that gcd(b, c) = d with b = dB and c = dC for B
and C coprime.

Then
  
    B  C
b c dB dC d d
gcd a − 1, a − 1 = gcd a − 1, a − 1 = gcd a − 1, a − 1 = ad − 1,

by applying the above result.

So if d = 1, the effect of b and c disappears.
But d > 1, asking that gcd ab − 1, bc − 1 = 100 would be impossible because 101 is not a

power.

Also solved by Hatef I. Arshagi, Guilford Technical Community College, Jamestown,

NC; Kee-Wai Lau, Hong Kong, China, and the proposer.

25
• 5606: Proposed by Ovidiu Furdui and Alina Sîntămărian, Technical University of Cluj-Napoca,
Cluj-Napoca, Romania
Let a, b > 0, c ≥ 0 and 4ab − c2 > 0. Calculate
Z ∞
x
dx.
−∞ aex + be−x + c

Solution 1 by Seán M. Stewart, Bomaderry, NSW, Australia

Denote the integral to be evaluated by I(a, b, c) where a, b > 0 and c ≥ 0 such that 4ab − c2 > 0.
We shall show that √ !
log(b/a) 4ab − c2
I(a, b, c) = √ arctan .
4ab − c2 c
Writing the integral as
∞
xex
Z
I(a, b, c) = dx,
−∞ ae2x + cex + b
enforcing a substitution of x 7→ log(x) produces
Z ∞
log(x)
I(a, b, c) = 2
dx.
0 ax + cx + b
q
Since a, b > 0, letting x = t ab yields
q 
r Z log t ab
b ∞
I(a, b, c) = q dt
a 0 bt2 + ct b + b
a
r  Z ∞
1 b b dt
= log q
2 a a 0 bt2 + ct ab + b
r Z
b ∞ log(t)
+ q dt. (1)
a 0 bt2 + ct b + b
a

Enforcing a substitution of t 7→ 1t in the second of the integrals after the equality in (1)
immediately shows it has a value equal to zero. Thus
r  Z ∞
1 b b dt
I(a, b, c) = log q . (2)
2 a a 0 2
bt + ct b
+b a

Completing the square in the denominator of the integrand given in (2) one has
 Z ∞
1 b dt
I(a, b, c) = √ log 2  . (3)
a

2 ab 0 t + 2√cab + 4ab−c
2
4ab

4ab−c2
The constant term 4ab appearing in the denominator of the integrand of (3) is positive since

26
4ab − c2 > 0 and a, b > 0. Performing the integration, which is elementary, we have
" √ !#∞
log(b/a) 2t ab + c
I(a, b, c) = √ arctan √
4ab − c2 4ab − c2 0
  
log(b/a) π c
=√ − arctan √
4ab − c2 2 4ab − c2
√ !
log(b/a) 4ab − c2
=√ arctan ,
4ab − c2 c

as announced. Note in the last line we have made use of the following well-known identity for
the arctangent function of
 
1 π
arctan(x) + arctan = , x > 0,
x 2

Solution 2 by Ulrich Abel, Technische Hochschule Mittelhessen, Germany

We show that for a, b > 0 and c ≥ 0
q √ 
b √4ab−c


Z ∞
x 2 log a arctan 4ab+c
I := dx = √ .
−∞ aex + be−x + c 4ab − c2

The change of variable x = t + 12 log ab leads to

∞ t + 12 log ab
Z
I= √ dt.
−∞ ab (et + e−t ) + c
t
Since √
ab(et +e−t )+c
is an odd integrable function, we obtain
 Z ∞
1 b 1
I = log √ dt
2 a −∞ ab (e + e−t ) + c
t
 Z ∞
1 b 1
= √ log dt,
4 ab a −∞ cosh t + g
p
where g = c2 / (4ab) < 1 by assumption. Since
r 
1−g
Z
p
2
1 t
1−g dt = 2 arctan tanh
cosh t + g 1+g 2
we have
∞ r 
1−g
Z
p 1
1 − g2 dt = 4 arctan ,
−∞ cosh t + g 1+g
which implies the desired formula.
Remark: In the particular case c = 0, we obtain g = 0 and
Z ∞  
x π b
x + be−x
dx = √ log .
−∞ ae 4 ab a

Solution 3 by Kee-Wai Lau, Hong Kong, China

27
Denote the given Integral by I. We show that
 
c
(ln b − ln a) cos−1 √
√ 2 ab
I= (1)
4ab − c2
Z ∞
1 ln y
By the substitution x = ln y, we obtain I = dy. It is known
0 a + cy y2
a + a
b

([1], p.537, entry 4.233(5)) that for k > 0 and 0 < t < π, we have
Z ∞
ln x t ln k
2 2
dx = .
0 x + 2xk cos t + k k sin t
r  
b −1 c
By putting k = and t = cos √ , we obtain (1) readily.
a 2 ab
1. I.S. Gradshteyn and I.M. Ryzhik. Table of Integrals, Series, and Products,
Seventh Edition, Elsevier, Inc. 2007.

Solution 4 by Peter Fulop, Gyomro, Hungary

Let a, b > 0 and 4ab − c2 ≥ 0.

R∞ x
Calculate I = dx
−∞ aex + be−x + c

Partial fraction decomposition

Let’s transform the integral into two parts:

Z∞ Z0
x x
I= dx − dx (15)
ae + b/ex + c
x aex + b/ex + c
0 −∞

At first appling the x → −x substitution for the second integral of the (1) and then the
substitution of t = e1x for both integrals of (1):

Z1 Z1
1 ln(t) 1 ln(t)
I= dt − dt (16)
a 2 c b
t + at + a b t2 + cb t + ab
0 0

Find the roots of the quadratic expressions of the (2):

√ r
−c ± i 4ab − c2 b ∓iϕ
t1,2 = = e (17)
2a a

√ r
−c ± i 4ab − c2 a ∓iϕ
t3,4 = = e (18)
2b b
√
where ϕ = arctan( dc ), d = 4ab − c2 and i2 = −1 the (2) is becoming the following:

28
 Z1  1  Z1
1 ln(t) ln(t) 
I= dt − dt (19)
a (t − t1 )(t − t2 ) b (t − t3 )(t − t4 )
0 0

Taking into account h that: i

1 a 1 1
(t−t1 )(t−t2 ) = id − (t−t1 ) + (t−t2 ) and
h i
1 b 1 1
(t−t3 )(t−t4 ) = id − (t−t3 ) + (t−t4 ) the (5) will be the following:

Z1 Z1 Z1 Z1
i ln(t) ln(t) ln(t) ln(t) 
I= − dt + dt + dt − dt (20)
d (t − t1 ) (t − t2 ) (t − t3 ) (t − t4 )
0 0 0 0

Spence function and its properties

Based on (6) let Id
i = −I1 + −I2 + I3 − I4 respect to tk roots (k=1,2,3,4). Let’s pull out −tk
from all denominators and performing the substitutions x = ttk .

1 1
Ztk Ztk
ln(x) + ln(tk ) ln(x) ln(tk )
Ik = dx = + dx (21)
(1 − x) 1 − x (1 − x)
0 0

1
Ztk
ln(x) 1 1
Ik = dx + ln( ) ln(1 − ) (22)
1−x tk tk
0

Introducing further substitution (r = 1 − x) regarding the integral of (8) we get:

1− t1
Z1 k
ln(1 − r) ln(1 − r)
Z
1 1
Ik = dr − dx + ln( ) ln(1 − ) (23)
r r tk tk
0 0

Using the definition of the Spence function (Dilogaritm function) we have:

1 1 1
Ik = −Li2 (1) + Li2 (1 − ) + ln( ) ln(1 − ) (24)
tk tk tk
Applying the following identity of the Spence function:
Li2 (z) + Li2 (1 − z) = Li2 (1) − ln(z) ln(1 − z), (10) will be the following:

1
Ik = −Li2 ( ) (25)
tk
1 1
Based on (3),(4) can be seen that t4 = t1 and t3 = t2 , go back to (6) we get value of the
integral (I):

29
 1 1
I = Li2 ( ) − Li2 ( ) − Li2 (t2 ) + Li2 (t1 ) (26)
t1 t2

Using the following identity Li2 (z) + Li2 ( z1 ) = −Li2 (1) − 12 ln2 (−z) twice we get:

i t2
I= ln( ) ln(t1 t2 ) (27)
2d t1
Finally substitute back t1 and t2 from (3) we get the result:

ln( a )
q
I= √ b
4ab−c2
arctan( 4ab
c − 1) (28)

Also solved by Albert Stadler, Herrliberg, Switzerland, and the proposers.

30