68 Normat 56:2, 68–79 (2008)
Integer Crossed Ladders; parametric represen-
tations and minimal integer values
Ralph Høibakk
a
, Tron Jorstad
c
Dag Lukkassen
b
and Lars-Petter Lystad
a
a
Narvik University College
b
Narvik University College and Norut Narvik
c
Apartado 1, 35580 Playa Blaca, Lanzarote, Canary islands, Spain
1 Introduction
The so-called ladder problems have attracted the interest of mathematicians through
centuries. For an overview and a historical bibliography, see [3] and [4]. One such
problem is the crossed ladders problem (CLP), which can be formulated as follows:
Two ladders of length a and b lean against two vertical walls. The ladders cross
each other at a point with distance c above the floor (see Figure 1). Determine
the distance x between the walls and the height above the floor y and z where the
ladders touch the walls.
Figure 1: The ladder problem.
The existence of integer solutions
of this problem is well known (see
[1], [6], [8] and [7]). However, the ac-
tual calculation of integer solutions
using known methods is cumber-
some. In this paper we introduce es-
sentially simpler methods for calcu-
lating all possible integer solutions
by means of four parameters. We
also identify several interesting sub-
classes of CLP-solutions described
by fewer parameters or obtained re-
cursively by using sequences of Pell
type. A study of minimal integer
values is also included. Minimal in-
teger values have been discussed
earlier in [6] and [7]. In this paper
we formulate new methods for iden-
tifying such values and present val-
ues smaller than the ones formerly
published.
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 69
The paper is organized as follows. In Section 2 we give some remarks concerning
Pythagorean triples and the trivial case a = b, y = z.Parametric representations of
general solutions are collected in Section 3. An alternative method for parametric
representation of several interesting subclasses can be found in Section 4. In Section
5 and Section 6 we study minimal integer values, in particular the minimal value
of x, which appears in a certain subclass of CLP-solutions. The connection be-
tween this subclass and recursive formulae of type Pell numbers is found in Section
7.Similarly, in Section 8 we describe some subclasses of CLP-solutions containing
minimum values for c and z.
2 On Pythagorean triples
By the above formulation of the crossed ladders problem we have that
(
x
2
= a
2
− y
2
= b
2
− z
2
,
c =
yz
y + z
.
(1)
The case a = b, y = z is trivial, since the set of all Pythagorean triples (x, a, y)
(i.e. triples of positive integers satisfying x
2
= a
2
− y
2
) is precisely those triples of
the forms
(x, a, y) =
m
2
− n
2
, m
2
+ n
2
, 2mn
,(2)
(x, a, y) =
2mn, m
2
+ n
2
, m
2
− n
2
,(3)
where m = M
√
T , n = N
√
T , for some positive integers M, N and T . This fact is
well known when n, m are integers and gcd(x, a, y) = 1 (see [2]). The general case
follows directly from this result by putting T = gcd(x, a, y). Since c = y/2, (2) will
always give a solution to (1) and (3) will give a solution if 2 | m
2
− n
2
. Therefore,
in the rest of this paper we will consider the non-trivial case assuming that
a > b (or y > z). (4)
Note also that if x
2
= a
2
− y
2
= b
2
− z
2
, then it is clear that we obtain an
integer solution to (1) by replacing (a, b, c, x, y, z) with SG
−1
(a, b, c, x, y, z), where
G = gcd(a, b, c, x, y, z) and S is the scaling parameter:
S =
y + z
gcd(yz, y + z)
.
We also observe that if
r = 2m
1
n
1
, s = m
2
1
− n
2
1
, t = m
2
1
+ n
2
1
, (5)
then
2r = m
2
2
− n
2
2
, 2s = 2m
2
n
2
, 2t = m
2
2
+ n
2
2
,
70 Høibakk, Jorstad, Lukkassen and Lystad Normat 2/2008
where m
2
= m
1
+ n
1
, n
2
= m
1
− n
1
. Moreover, replacing m
1
and n
1
with
√
2m
1
and
√
2n
1
in (5) we see that we also might put
2r = 2m
1
n
1
, 2s = m
2
1
− n
2
1
, 2t = m
2
1
+ n
2
1
.
Remark 1. This shows that for any Pythagorean triple (t, r, s), satisfying t
2
=
r
2
+ s
2
, we have that 2(t, r, s) can be represented in both forms
2(t, r, s) =
m
2
+ n
2
, 2mn, m
2
− n
2
,(6)
2(t, r, s) =
m
2
+ n
2
, m
2
− n
2
, 2mn
,(7)
where m = M
√
T , n = N
√
T , for some positive integers M, N and T . As we will
see in the following sections, this implies that we in principal may use only one of
these representations as a tool for finding CLP-solutions.
Remark 2. Letting m and n be integer values only, we are still able to represent
any primitive Pythagorean triple (t, r, s) (satisfying t
2
= r
2
+ s
2
, gcd(r, s, t) = 1),
or its double (2t, 2r, 2s) in both forms (m
2
+ n
2
, 2mn, m
2
−n
2
) and (m
2
+ n
2
, m
2
−
n
2
, 2mn).
3 Parametric representations of general solutions
Lemma 3. Suppose that a, y, b, z, x and c are positive integers such that (1) holds.
Then,
(a, y, b, z, x) = Sw
−1
(t
1
s
2
, r
1
s
2
, t
2
s
1
, r
2
s
1
, s
1
s
2
) (8)
for some positive integers t
i
, r
i
, and s
i
satisfying the conditions
t
2
i
− r
2
i
= s
2
i
,
w = gcd(s
1
, s
2
),
S =
r
1
s
2
+ r
2
s
1
gcd(r
1
s
2
r
2
s
1
, r
1
s
2
+ r
2
s
1
)
.
(9)
Conversely, any set of integers of the form (8) satisfying (9) also satisfies (1).
Proof. Let s
0
2
= gcd(a, y) and s
0
1
= gcd(b, z). Let t
1
, r
1
, t
2
and r
2
be positive
integers such that a = t
1
s
0
2
, y = r
1
s
0
2
, b = t
2
s
0
1
and z = r
2
s
0
1
, where gcd(s
0
1
, s
0
2
) = 1.
By (1),
x
2
=
t
2
1
− r
2
1
s
02
2
=
t
2
2
− r
2
2
s
02
1
.
Thus, we find that
t
2
i
− r
2
i
= s
02
i
w
2
, i = 1, 2, (10)
for some integer w. By putting s
i
= s
0
i
w we obtain (9). Conversely, if (9) holds,
then it is seen directly by inspection that (1) holds. This completes the proof.
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 71
Since (t
i
, r
i
, s
i
) of Lemma 3 is a Pythagorean triple, we know that (t
i
, r
i
, s
i
) =
(t
i
, u
i
, v
i
) or (t
i
, r
i
, s
i
) = (t
i
, v
i
, u
i
), where
t
i
= m
2
i
+ n
2
i
, u
i
= 2m
i
n
i
, v
i
= m
2
i
− n
2
i
,
m
i
= M
i
p
S
i
, n
i
= N
i
p
S
i
,
for some positive integers M
i
, S
i
and N
i
. Conversely, it is seen directly that
(t
i
, r
i
, s
i
) is a Pythagorean triple for any positive integers m
i
, n
i
. This leaves us
with four alternative representations of the solution vector
(a, y, b, z, x) = Sw
−1
(t
1
s
2
, r
1
s
2
, t
2
s
1
, r
2
s
1
, s
1
s
2
), (11)
from which we are able to reproduce all possible solutions of the crossed ladders
problem in terms of the integer parameters m
1
, m
2
, n
1
and n
2
. The four alternative
representations are obtained by putting (t
i
, r
i
, s
i
) equal to the following values:
1. (t
1
, r
1
, s
1
) = (t
1
, u
1
, v
1
) and (t
2
, r
2
, s
2
) = (t
2
, u
2
, v
2
),
2. (t
1
, r
1
, s
1
) = (t
1
, u
1
, v
1
) and (t
2
, r
2
, s
2
) = (t
2
, v
2
, u
2
),
3. (t
1
, r
1
, s
1
) = (t
1
, v
1
, u
1
) and (t
2
, r
2
, s
2
) = (t
2
, u
2
, v
2
),
4. (t
1
, r
1
, s
1
) = (t
1
, v
1
, u
1
) and (t
2
, r
2
, s
2
) = (t
2
, v
2
, u
2
).
For example, for the latter case (case 4) we obtain
a = Sm
2
n
2
(m
2
1
+ n
2
1
), (12)
b = Sm
1
n
1
(m
2
2
+ n
2
2
), (13)
c =
m
1
n
1
m
2
n
2
(m
2
1
− n
2
1
)(m
2
2
− n
2
2
)
gcd
1
, (14)
x = S2m
1
n
1
m
2
n
2
, (15)
y = Sm
2
n
2
(m
2
1
− n
2
1
), (16)
z = Sm
1
n
1
(m
2
2
− n
2
2
), (17)
where
S =
m
2
n
2
(m
2
1
− n
2
1
) + m
1
n
1
(m
2
2
− n
2
2
)
gcd
1
,
gcd
1
= gcd((m
1
n
1
m
2
n
2
(m
2
1
− n
2
1
)(m
2
2
− n
2
2
), m
2
n
2
(m
2
1
− n
2
1
) + m
1
n
1
(m
2
2
− n
2
2
)).
Note also that by (4) we may assume that m
2
n
2
(m
2
1
+ n
2
1
) > m
1
n
1
(m
2
2
+ n
2
2
) (or
equivalently that m
1
/n
1
> m
2
/n
2
).
Remark 4. Replacing (t
i
, r
i
, s
i
) with (2t
i
, 2r
i
, 2s
i
) will change (a, y, c, b, z, x) ob-
tained by (8) to (2a, 2y, 2c, 2b, 2z, 2x). Hence, in view of Remark 1 the represen-
tation (12)-(17) will at least give us all even CLP-solutions. Thus, including the
even solutions obtained from (12)-(17) and subsequently divided by 2, we obtain the
complete set of CLP-solutions.
72 Høibakk, Jorstad, Lukkassen and Lystad Normat 2/2008
4 Alternative representations
If
x
2
= a
2
− y
2
= b
2
− z
2
, (18)
then (x, a, y) and (x, b, z) can be represented as Pythagorean triples in four alter-
native ways, for example as
(x, a, y) =
p
2
− q
2
, p
2
+ q
2
, 2pq
,(19)
(x, b, z) =
r
2
− s
2
, r
2
+ s
2
, 2rs
,(20)
p
2
− q
2
= r
2
− s
2
.(21)
Hence, we obtain an integer solution to (1) by replacing (a, b, c, x, y, z), c = yz/(y +
z), with S(a, b, c, x, y, z), where S is the scaling parameter:
S =
pq + rs
gcd(pqrs, pq + rs)
.
This representation may be used to identify certain subclasses of solutions to (1).
For example, observing that the identities
(2N
2
1
− N
2
2
)
2
−
2N
2
1
2
=
N
2
2
2
− (2N
1
N
2
)
2
, (22)
(N
2
1
− 2N
2
2
)
2
−
N
2
1
2
= (2N
2
)
2
− (2N
1
N
2
)
2
, (23)
are on the form (21), we obtain solutions by putting (p, q, r, s) = (2N
2
1
−N
2
2
, 2N
2
1
, N
2
2
, 2N
1
N
2
)
or (p, q, r, s) = (N
2
1
− 2N
2
2
, N
2
1
, 2N
2
, 2N
1
N
2
).
Furthermore, by using the identities
(N
2
1
± N
2
2
)
4
= N
8
1
± 4N
6
1
N
2
2
+ 6N
4
1
N
4
2
± 4N
2
1
N
6
2
+ N
8
2
,
(N
4
1
± N
4
2
)
2
= N
8
1
± 2N
4
1
N
4
2
+ N
8
2
,
(N
3
1
N
2
± N
1
N
3
2
)
2
= N
6
1
N
2
2
± 2N
4
1
N
4
2
+ N
2
1
N
6
2
,
we can obtain the following identities
4N
4
1
N
4
2
= (N
4
1
+N
4
2
)
2
−(N
4
1
−N
4
2
)
2
= (N
3
1
N
2
+N
1
N
3
2
)
2
−(N
3
1
N
2
−N
1
N
3
2
)
2
, (24)
(N
4
1
− N
4
2
)
2
= (N
2
1
+ N
2
2
)
4
− 4(N
3
1
N
2
+ N
1
N
3
2
)
2
= (N
4
1
+ N
4
2
)
2
− 4N
4
1
N
4
2
, (25)
4(N
3
1
N
2
− N
1
N
3
2
)
2
= 4(N
3
1
N
2
+ N
1
N
3
2
)
2
− 16N
4
1
N
4
2
= (N
4
1
− N
4
2
)
2
− (N
2
1
− N
2
2
)
4
.
(26)
By noting that each of the identities (24), (25) and (26) is on the form (18) we may
obtain additional representations of subclasses of solutions for the crossed ladders
problem. For example, by using (24) we see directly that
(x, a, y, b, z) =
2N
2
1
N
2
2
, N
4
1
+ N
4
2
, N
4
1
− N
4
2
, N
1
N
2
(N
2
1
+ N
2
2
), N
1
N
2
(N
2
1
− N
2
2
)
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 73
is a solution to (18) for any integers N
1
, N
2
. Hence, replacing (a, b, c, x, y, z) with
S(a, b, c, x, y, z) (as before c = yz/(y + z)) where S is the scaling parameter
S =
y + z
gcd(yz, y + z)
= N
2
1
+ N
1
N
2
+ N
2
2
,
we obtain one type of solution for the crossed ladders problem, namely the one of
the form
(x, a, y, b, z) = S
2N
2
1
N
2
2
, N
4
1
+ N
4
2
, N
4
1
− N
4
2
, N
1
N
2
(N
2
1
+ N
2
2
), N
1
N
2
(N
2
1
− N
2
2
)
,
(27)
c = N
1
N
2
(N
4
1
− N
4
2
). (28)
5 On minimum solutions
Minimum solutions has been considered earlier in [6] and [7]. However, the search
for such quantities has mostly been based on tabulating solutions for increasing
values of x. In this paper we will attack this problem from a different angle. The
main idea is to represent sufficiently rich classes of solutions in such a way that
minimum values can be identified more directly.
In [7] the following values are described as ”the smallest possible solution”:
x = 56, a = 119, b = 70, c = 30, y = 105, z = 42. By our methods this solution is
obtained by putting m
1
= 4, n
1
= 1, m
2
= 2, n
2
= 1 in the formulae (12-17), or
N
1
= 2, N
2
= 1 in formulae (27) and (28).
It is not clear what is meant by ”smallest possible solution” in the article [7].
By our investigation the above solution appears to have the property that the sum
x + a + b + c + y + z becomes smallest possible among all solutions. However, as we
will show in this paper, none of the individual quantities x, a, b, c, y, z is minimal
for this solution.
6 Minimum value for x
For the purpose of finding minimum values it is suitable to choose the parameters
m
i
and n
i
such that scaling factor S = 1. In this way we avoid the calculation of S,
which often is cumbersome. But more important, any subsequent scaling without
knowing the value of S in advance, makes it hard to choose parameters such that
the value of x gets minimal.
By (1) we have that gcd(yz, y + z) = y + z. Thus, z and y can be represented as
follows,
y = m(m + n), z = n(m + n), (29)
where m = p
√
t and n = q
√
t for some integers p, q, t such that p > q. In order to
verify this representation, we put y = pk and z = qk, where k = gcd(y, z). Hence,
gcd(p, q) = 1. (30)
74 Høibakk, Jorstad, Lukkassen and Lystad Normat 2/2008
Letting p + q = dτ, pq = ds, where d = gcd(p + q, pq), we find that p = d
1
s
1
,
q = d
2
s
2
for some integers d
1
, d
2
and s
1
, s
2
, satisfying d = d
1
d
2
and s = s
1
s
2
.
Hence, q = p + q − p = d
1
d
2
τ − d
1
s
1
= d
1
(d
2
τ −s
1
). Thus, since p = d
1
s
1
and
q = d
1
(d
2
τ −s
1
) , (30) gives that d
1
= 1. Similarly, we have that p = p + q − q =
d
1
d
2
τ − d
2
s
2
= d
2
(d
1
τ −s
2
) , i.e. p = d
2
(d
1
τ −s
2
) and q = d
2
s
2
, which, again
according to (30), gives that d
2
= 1. Thus
gcd(p + q, pq) = d = d
1
d
2
= 1. (31)
Moreover, since gcd(yz, y + z) = y + z, we have that yz = w(y + z) for some
integer w, i.e. pqk
2
= wk(p + q). Thus, pqk = w(p + q), and by (31), this gives that
k = t(p + q) for some integer t. Hence, y = pk = tp(p + q) and z = qk = tq(p + q).
Putting m = p
√
t and n = q
√
t we therefore obtain (29).
A different way of showing (29) is by putting
y = c + r, z = c + s (32)
for some integers r and s. From the identity c = yz/(y + z) we obtain the simple
relation
c
2
= rs. (33)
Next, we put r = tr
1
, s = ts
1
, where t = gcd(r, s) (such that gcd(r
1
, s
1
) = 1).
By (33) we see that r
1
s
1
is a perfect square, and since r
1
and s
1
do not have any
prime factors in common, r
1
and s
1
have to be perfect squares, i.e. on the forms
r
1
= p
2
, s
1
= q
2
. Thus
y = c + r =
√
rs + r = p
√
t(q
√
t + p
√
t) = m(m + n), (34)
z = c + s =
√
rs + s = q
√
t(p
√
t + q
√
t) = n(m + n). (35)
According to (19), (20), (21) and (29), one family of solutions to (1) is parametrized
as follows:
x = 2N
1
N
2
= 2N
3
N
4
, (36)
y = N
2
1
− N
2
2
= m(m + n), (37)
z = N
2
3
− N
2
4
= n(m + n), (38)
a = N
2
1
+ N
2
2
, b = N
2
3
+ N
2
4
, c = mn. (39)
In addition, we restrict ourselves to the case
N
1
+ N
2
= m + n, N
1
− N
2
= m, (40)
N
3
= KM
3
, N
4
= KM
4
, K
2
= n, (41)
M
2
3
− M
2
4
= m + n, (42)
n = 2n
1
. (43)
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 75
Thus, we obtain that
m = 2M
3
M
4
− n
1
, (44)
N
1
= m + n
1
= 2M
3
M
4
, N
2
= n
1
, N
3
= M
3
√
2n
1
, (45)
N
4
= M
4
√
2n
1
. (46)
According to (42), (43) and (44) we have that
M
2
3
− M
2
4
− 2M
3
M
4
− n
1
= 0 (47)
i.e.
M
3
= M
4
+
q
2M
2
4
+ n
1
. (48)
Now, choosing M
3
, M
4
and n
1
such that (48) is satisfied, we directly obtain solu-
tions of (1) without performing any subsequent scaling. However, for the purpose
of finding minimal solutions, it may be necessary to divide the obtained solution
(a, b, c, x, y, z) with gcd(a, b, c, x, y, z). By (36) and (45) we obtain x = 4n
1
M
3
M
4
.
For each fixed n
1
, the smallest integers M
3
> 0, M
4
> 0 satisfying (48) are de-
termined by the smallest integer M
4
> 0 making 2M
2
4
+ n
1
a perfect square. By
inspection it is easy to see that x is smallest possible when n
1
= 2, M
3
= 1 and
M
4
= 3. However, these values makes y = z, the trivial case, which contradicts
our preliminary assumption (4). Except for this case, we observe that x is smallest
possible when n
1
= 1, M
3
= 2 and M
4
= 5, giving the values
N
1
= 20, N
2
= 1, N
3
= 5
√
2; N
4
= 2
√
2, m = 19, n = 2,
which gives the solution
x = 40, a = 401, b = 58, c = 38, y = 399, z = 42.
This appears to be the solution which gives the smallest integer value of x among
all possible solutions satisfying (4). In fact, by factorizing each integer between 1
2
and 39
2
and using that x
2
= (a − y) (a + y) = (b − z) (b + z) it is possible to verify
that there exists no integer solution for x < 40 such that gcd(yz, y + z) = y + z.
7 Connection to Pell numbers
For n
1
= 1, we find an interesting connection between the class of solutions satisfy-
ing the condition (48) and the well-known Pell numbers {P
i
}
∞
i=0
, which are defined
recursively by
P
i
= 2P
i−1
+ P
i−2
, P
0
= 0, P
1
= 1. (49)
This sequence has the property that if (x, y) = (P
2i
+ P
2i−1
, P
2i
) then
x =
p
2y
2
+ 1, (50)
76 Høibakk, Jorstad, Lukkassen and Lystad Normat 2/2008
i.e. (x, y) is a solution to Pell’s Equation x
2
− 2y
2
= 1. Concerning this and other
basic properties of Pell numbers we refer to the literature (see e.g. [5] and the
references given therein). Adding P
2i
to (50) and using (49) we obtain that P
2i+1
=
P
2i
+
p
2P
2
2i
+ 1. Hence, comparing with (48) we obtain a class of solutions by
putting
M
3
= P
2i+1
, M
4
= P
2i
, m = 2P
2i
P
2i+1
− 1, n = 2, (51)
N
1
= 2P
2i
P
2i+1
, N
2
= 1, N
3
= P
2i+1
√
2, N
4
= P
2i
√
2,
which gives the following integer solution to our problem:
a = 4P
2
2i
P
2
2i+1
+ 1, b = 2(P
2
2i+1
+ P
2
2i
), c = 2(2P
2i
P
2i+1
− 1), (52)
x = 4P
2i
P
2i+1
, y = 4P
2
2i
P
2
2i+1
− 1, z = 2(P
2
2i+1
− P
2
2i
). (53)
For the calculation of the Pell numbers, the following explicit formula may be useful
P
i
=
(1 +
√
2)
i
− (1 −
√
2)
i
2
√
2
.
We also observe that b+z, b−z and b
2
−z
2
are perfect squares equal to (2P
2i+1
)
2
, (2P
2i
)
2
and (4P
2i
P
2i+1
)
2
, respectively.
By choosing different values of n
1
we obtain other families of integer solutions
of (1) generated by pairs (M
3
, M
4
) of the type M
3
= P
k
, M
4
= P
l
. For example,
if n
1
= 2 we obtain the same definition of the sequence P
i
as above except that
P
0
= 1, i.e.
P
i
= 2P
i−1
+ P
i−2
, P
0
= 1, P
1
= 1, (54)
with corresponding explicit formula
P
i
=
(1 +
√
2)
i
+ (1 −
√
2)
i
2
.
In this case we put M
3
= P
2i
, M
4
= P
2i−1
. All obtained solutions for n
1
= 2 may
be divided by 8.
The minimal non-trivial solution in the subclass generated from the case n
1
= 2
is
x = 119, a = 7081, b = 169, c = 118, y = 7080, z = 120.
As a curiosity we want to add that this appears to be the smallest possible solution
satisfying a − y = 1. In addition, this also appears to be the smallest possible
solution where three elements have consecutive values like c = 118, x = 119, z =
120. It is possible to show that all solutions generated from the case n
1
= 2 (after
division by the greatest common divisor) have the property a − y = 1, z − x = 1
and c, x, z have consecutive values. As for the case n
1
= 1, b + z, b −z and b
2
−z
2
are perfect squares, equal to (P
2i
)
2
, (P
2i−1
)
2
and (P
2i
P
2i+1
)
2
, respectively.
It can be shown that there exists no integers M
3
and M
4
satisfying (48) for
n
1
= 5. For n
1
= 7 we have that M
3
= P
2i+1
, M
4
= P
2i
, where {P
i
} are the
following Pell type of numbers,
P
2i+1
= 2P
2i
+ P
2i−3
, P
2i
= 2P
2i−3
+ P
2i−4
, P
0
= 0, P
1
= 1.
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 77
For n
1
= u
2
, where u is an integer, M
3
= P
2i+1
, M
4
= P
2i
, where {P
i
} is given by
P
i
= 2P
i−1
+ P
i−2
, P
0
= 0, P
1
= u.
8 Minimum values for c and z
It is known that 14 and 15 are the minimum values for c and z, respectively.
Similarly as above we will in this section develop a representation of a sufficiently
rich class of solutions, from which these minimal solutions are derived more directly
than the cumbersome tabulation approach.
It is reasonable to assume that the minimum values for c and z appear when
the differences z − c and b − x are small (see Figure 1). Recall that by putting
y − c = r, (55)
z − c = s,
we obtain from the identity c = yz/(y + z) the simple relation
c
2
= rs. (56)
We restrict ourselves to the case
z − c = s = 1, b − x = 1. (57)
Let x = x
1
+ x
2
, where x
1
denotes the horizontal distance between the left wall
and the intersection point. By comparing equilateral triangles we obtain the re-
lations x
1
/ (z − c) = x
1
= x
2
/c (which implies that x = x
1
(1 + c)), and b/x =
q
(z − c)
2
+ x
2
1
/x
1
, i.e.
x + 1
x
=
x
1
(1 + c) + 1
x
1
(1 + c)
=
p
1 + x
2
1
x
1
. (58)
Solving (58) we find that
x =
1
2
c (2 + c) . (59)
By (55), (56), (57) and (59) we have that
y = c
2
+ c, z = 1 + c, b =
1
2
c (2 + c) + 1, (60)
and, hence, by the formula a
2
= x
2
+ y
2
, we find that
78 Høibakk, Jorstad, Lukkassen and Lystad Normat 2/2008
a
2
=
5
4
c
4
+ 3c
3
+ 2c
2
,
which shows that c is even, i.e. of the form c = 2p. Accordingly,
a
2
= 4p
2
(5p
2
+ 6p + 2), (61)
b = 2p
2
+ 2p + 1, c = 2p, (62)
x = 2p(p + 1), y = 2p(2p + 1), z = 2p + 1. (63)
Due to (61) we have that 5p
2
+ 6p + 2 is a perfect square. Besides for p = −1
(which we ignore since it represents the trivial case), the smallest positive value
of c = 2p satisfying this condition is obtained for p = 7. This value gives us the
following integer solution
a = 238, b = 113, c = 14, x = 112, y = 210, z = 15, s = 1, r = 196,
for which both z and c are minimal. We also note that this solution also gives the
minimum values for b − x and z − c.
A sequence of integer values of p satisfying the above condition can be found by
putting
5p
2
+ 6p + 2 = (2p + q)
2
(64)
for some integer q, which yields the following explicit expressions
p = 2q − 3 ±
p
5q
2
− 12q + 7, (65)
q = −2p ±
p
5p
2
+ 6p + 2. (66)
We note that if one of the values given by (65) is an integer (for a fixed integer value
of q), so is the other one. The same holds for (66). This explains why we choose
to express the right side of (64)) on the form (2p + q)
2
. We may now produce an
infinite family of pairs (p, q) = (p
2i+1
, q
2i
) satisfying the identity (64) by using the
recursive formulae
p
2i+1
= 2q
2i
− 3 +
q
5q
2
2i
− 12q
2i
+ 7, q
0
= 1, (67)
q
2i
= −2p
2i−1
+
q
5p
2
2i−1
+ 6p
2i−1
+ 2. (68)
From these formulae we obtain the following values:
q
0
= 1 → p
1
= −1 → q
2
= 3 → p
3
= 7 →
q
4
= −31 → p
5
= −137 → q
6
= 579 → p
7
= 2447 → ... .
Normat 2/2008 Høibakk, Jorstad, Lukkassen and Lystad 79
Note that by inserting any value p = p
i
, obtained in this way, into (61), (62) and
(63) we obtain solutions of our problem satisfying the condition (57). Interestingly,
all these solutions have the property that
y/z = c, (69)
which follows directly by (60).
We also want to add that the above sequences also can be obtained by the
following recursive formulae
q
2i
= −4p
2i−1
− q
2i−2
, (70)
p
2i+1
= 4q
2i
− p
2i−1
− 6, (71)
q
0
= 1, p
1
= −1. (72)
When the condition (57) is satisfied we obtain from (69) that F ≡ y/z is an integer.
For the class corresponding to n
1
= 1 in Section 7, we obtain from (37), (38) and
(51) that
F =
y
z
=
m
n
=
2P
2i
P
2i+1
− 1
2
, (73)
which never is an integer. On the other hand, for the class corresponding to n
1
= 2,
F =
y
z
=
m
n
=
2P
2i
P
2i−1
− 2
4
=
P
2i
P
2i−1
− 1
2
. (74)
By induction it is directly seen from (54) that the sequence {P
i
} consists of odd
numbers. Thus, F = y/z is an integer according to (74).
References
[1] A. A. Bennett, E 433 (1940, 487), Amer. Math. Month., April (1941), 268-269.
[2] D. M. Burton, Elementary Number Theory, 5th ed. International Series in Pure
and Applied Mathematics. New York, McGraw-Hill (2002).
[3] K. Holing, På gjengrodde stiger, Normat 2 (1977), 62 - 78.
[4] K. Holing, På gjengrodde stiger - Epilog, Normat 2 (2002), 92-95.
[5] D. Kalman and R. Mena, The Fibonacci Numbers - Exposed. Mathematics Mag-
azine 76, 3 (2003).
[6] H. l. Nelson, The Two Ladders, J. Recreational Math. 11, 4 (1979-80) 312-314.
[7] A. Sutcliffe, Complete Solution of the Ladder Problem in Integers, Math. Gaz.
47 (1963), 133-136.
[8] H. E. Tester, Note 3036, Math. Gaz. 46(1962), p.313 .
