Solutions
Notes. A sequence x_{1}, x_{2}, ¼, x_{k} is in
arithmetic progression iff x_{i+1}  x_{i} is constant
for 1 £ i £ k1. A triangular number is a positive
integer of the form
T(x) º 
1
2

x(x+1) = 1 + 2 + ¼+ x , 

where x is a positive integer.
[ ¼] refers to the area of the figure between brackets.

332.

What is the minimum number of points that can be
found (a) in the plane,
(b) in space, such that each point in, respectively, (a) the plane,
(b) space, must be at an irrational distance from at least one of
them?
Solution 1. We solve the problem in space, as the planar problem
is subsumed in the spatial problem. Two points will never do, as we
can select on the right bisector of the segment joining them a point
that is the same rational distance from both of them (why?).
However, we can find three points that will serve. Select three
collinear points A, B, C such that AB  = BC  = u where u^{2} is not rational. Let P
be any point in space. If P, A, B, C are collinear and
PA  = a, PC  = c, then PB 
is equal to either a  u or c  u. If a and c are rational,
then both a  u and c  u are nonrational. Hence at least one
of the three distances is rational.
If P, A, C are not collinear, then PB is a median of
triangle PAC. Let b = PB . Then
a^{2} + c^{2} = 2b^{2} + 2u^{2} (why?). Since u^{2} is non rational,
at least one of a, b, c is nonrational.
Comment. You can check that a^{2} + c^{2} = 2(b^{2} + u^{2}) holds
for the collinear case as well.
Solution 2. [F. Barekat] As in the foregoing, the number has
to be at least three. Consider the points (0, 0, 0),
(u, 0, 0) and (v, 0, 0), where v is irrational and u and
v^{2} are rational. Let P ~ (x, y, z). Then the distances
from P to the three points are the respective square roots of
x^{2} + y^{2} + z^{2}, x^{2}  2ux + u^{2} + y^{2} + z^{2} and
x^{2}  2vx + v^{2} + y^{2} + z^{2}.
If the first of these is irrational, then we have one
irrational distance. Suppose that x^{2} + y^{2} + z^{2} is rational.
If x is irrational, then
x^{2}  2ux + u^{2} + y^{2} + z^{2} = (x^{2} + y^{2} + z^{2} + u^{2})  2ux 

is irrational. If x is rational, then
x^{2}  2vx + v^{2} + y^{2} + z^{2} = (x^{2} + y^{2} + z^{2} + v^{2})  2vx 

is irrational. Hence not all the three distances can be rational.

333.

Suppose that a, b, c are the sides of triangle
ABC and that a^{2}, b^{2}, c^{2} are in arithmetic progression.


(a) Prove that cotA, cotB, cotC are also
in arithmetic progression.


(b) Find an example of such a triangle where a, b, c are
integers.
Solution 1. [F. Barekat] (a) Suppose, without loss of generality
that a
£ b
£ c. Let AH be an altitude of the triangle.
Then
cotB = 
BH 
AH 

and cotC = 
CH 
AH 

. 

Therefore,
2[ABC] (cotB + cotC) = a AH  
æ è


BH + CH 
AH 


ö ø

= a^{2} . 

Similar equalities hold for b
^{2} and c
^{2}.
Therefore
2b^{2} = a^{2} + c^{2} Û2(cotA + cotC) = (cotB + cotC) + (cotB + cotA)Û cotA + cotC = 2cotB . 

The result follows from this.
(b) Observe that a^{2} + c^{2} = 2b^{2} if and only if
(c  a)^{2} + (c + a)^{2} = (2b)^{2}. So, if (x, y, z) is a
Pythagorean triple with z even and x and y of the
same parity, then
(a, b, c) = 
æ è


yx
2

, 
z
2

, 
y+x
2


ö ø

. 

Let (x, y, z) = (m
^{2}  n
^{2}, 2mn, m
^{2} + n
^{2}) where m and
n have the same parity and m > n. Then
(a, b, c) = 
æ è


n^{2} + 2mn  m^{2}
2

, 
m^{2} + n^{2}
2

, 
m^{2} + 2mn  n^{2}
2


ö ø

. 

To ensure that these are sides of a triangle, we need to impose
the additional conditions that
n^{2} + 2mn  m^{2} > 0 Û 2n^{2} > (m  n)^{2}Û m < (Ö2 + 1)n 

and
m^{2} + 2mn  n^{2} < (m^{2} + n^{2}) + (n^{2} + 2mn  m^{2})Û m < nÖ3 . 

Thus, we can achieve our goal as long as n
^{2} < m
^{2} < 3n
^{2} and
m
º n (mod 2).
For example, (m, n) = (5, 3) yields (a, b, c) = (7, 17, 23).
Comment. Take (x, y, z) = (2mn, m^{2}  n^{2}, m^{2} + n^{2})
to give the solution
(a, b, c) = 
æ è


m^{2}  2mn  n^{2}
2

, 
m^{2} + n^{2}
2

, 
m^{2} + 2mn  n^{2}
2


ö ø

. 

For example, (m, n) = (5, 1) yields (a, b, c) = (7, 13, 17).
Here are some further numerical examples; note how they come
in chains with the first of each triple equal to the last of
the preceding one:
(a, b, c) = [(1, 5, 7),] (7, 13, 17), (17, 25, 31), (31, 41, 49), ¼ 

(a, b, c) = (7, 17, 23), (23, 37, 47), (47, 65, 79), ¼ 

Solution 2. Since c^{2} = a^{2} + b^{2}  2ab cosC, we have
that
cotC = 
a^{2} + b^{2}  c^{2}
2absinC

= 
a^{2} + b^{2}  c^{2}
4[ABC]



with similar equations for cotA and cotB. Hence
cotA + cotC  2cotB = (2[ABC])^{1}(2b^{2}  a^{2}  c^{2}) . 

Thus, a
^{2}, b
^{2}, c
^{2} are in arithmetic progression if and only if
cotA, cotB, cotC are in arithmetic progression.
(b) We need to solve the Diophantine equation a^{2} + c^{2} = 2b^{2}
subject to the condition that c < a + b. The inequality is
equivalent to 2b^{2}  a^{2} < (a + b)^{2} which reduces to
(b  a)^{2} < 3a^{2} or b < (1 + Ö3)a. To ensure the
inequality, let us try b = 2a + k, so that
b^{2} = 4a^{2} + 4ka + k^{2} and we have to solve
7a^{2} + 8ka + 2k^{2} = c^{2}. Upon multiplication by 7 and shifting
terms, the equation becomes
(7a + 4k)^{2}  7c^{2} = 2k^{2} . 

(Note that b/a = 2 + (k/a) < 1 +
Ö3 as long as
a >
^{1}/
_{2}(
Ö3 + 1)k.)
We solve a Pell's Equation x^{2}  7y^{2} = 2k^{2} with
the condition that x º 4k (mod 7). There is a standard
technique for solving such equations. We find the fundamental
solution of x^{2}  7y^{2} = 1; this is the solution with the smallest
positive values of x and y, and in this case is (x, y) = (8, 3).
We need a particular solution of x^{2}  7y^{2} = 2k^{2}; the solution
(x, y) = (3k, k) will do. Then we get an infinite set of solutions
(x_{n}, y_{n}) for x^{2}  7y^{2} = 2k^{2} by defining (x_{0}, y_{0}) = (3k, k) and, for n ³ 1,
x_{n} + y_{n}Ö7 = (8 + 3Ö7)(x_{n1} + y_{n1}Ö7) . 

(Note that the same equation holds when we replace the plus signs
in the three terms by minus signs, so we can see that this works
by multiplying this equation by its surd conjugate.) Separating out
the terms, we get the recursion
x_{n} = 8x_{n1} + 21y_{n1} 

y_{n} = 3x_{n1} + 8y_{n1} 

for n
³ 1. Observe that x
_{n} º x
_{n1} (mod 7). Thus,
to get the solution we want, we need to select k such that
k
º 0 (mod 7).
An infinite family of solutions of x^{2}  7y^{2} = 2k^{2} starts with
(x, y) = (3k, k), (45k, 17k), (717k, 271k), ¼ . 

All but the first of these will yield a triangle, since we will
have 7a = x
 4k
³ 41k whence a > 5k >
^{1}/
_{2}(
Ö3+ 1)k. Let k = 7. Then, we get the triangles
(a, b, c) = (41, 89, 119), (713, 1433, 1897),
¼. We get
only similar triangles to these from other multiples of 7 for k.

334.

The vertices of a tetrahedron lie on the surface of
a sphere of radius 2. The length of five of the edges of the
tetrahedron is 3. Determine the length of the sixth edge.
Solution 1. Let ABCD be the tetrahedron with the lengths
of AB, AC, AD, BC and BD all equal to 3. The plane that
contains the edge AB and passes through the centre of the sphere is
a plane of symmetry for the tetrahedron and is thus orthogonal to
CD. This plane meets CD in P, the midpoint of CD.
Likewise, the plane orthogonal to AB passing
through the midpoint M of AB is a
plane of symmetry of the tetrahedron that passes through C, D and
P, as well as the centre O of the sphere.
Consider the triangle OCM with altitude CP. Since OM is the
altitude of the triangle OAB with OA  = OB  = 2 and AB  = 3, OM  = (Ö7)/2.
Since MC is an altitude of the equilateral triangle ABC,
MC  = (3Ö3)/2. Since OC is a radius of the
sphere, OC  = 2.
Let q = ÐOCM. Then, by the Cosine Law,
^{7}/_{4} = 4 + [ 27/4]  2Ö{27}cosq, so that
cosq = (Ö3)/2 and sinq = 1/2. Hence, the
area [OCM] of triangle OCM is equal to ^{1}/_{2}OC MC sinq = (3Ö3)/4. But this area is also
equal to ^{1}/_{2} CP OM  = ((Ö7)/4)CP . Therefore,
CD  = 2 CP  = 2(3Ö3)/(Ö7) = (6 Ö3)/(Ö7) = (6  Ö

21

)/7 . 

Comment. An alterntive way is to note that CMP is a
right triangle with hypotenuse CM with O a point on PM.
Let u = CP . We have that
CM  = (3Ö3)/2, OM  = (Ö7)/2, CO  = 2. so that
OP ^{2} = 4  u^{2} and MP  = [(Ö7)/2]+ Ö{4  u^{2}}. Hence, by Pythagoras' Theorem,

27
4

= 
é ë


Ö7
2

+  Ö

4  u^{2}


ù û

2

+ u^{2} 

Þ 
27
4

= 
7
4

+ Ö7  Ö

4  u^{2}

+ 4 

Þ u^{2} = 27/7 Þ u = (3Ö3)/Ö7 . 

Hence
CD
 = 2u = (6
Ö3)/
Ö7.
Solution 2. [F. Barekat] As in Solution 1, let CD be the
odd side, and let O be the centre of the sphere. Let G be the
centroid of the triangle ABC. Since the right bisecting
plane of the three
sides of triangle ABC each pass through the centroid G and the
centre O, OG ^ABC. Observe that CG  = Ö3
and GM  = (Ö3)/2, where M is the midpoint of
AB.
As CGO is a right triangle, GO ^{2} = CO ^{2}  CG ^{2} = 4  3 = 1, so that
GO  = 1. Hence, OM ^{2} = OG ^{2}+ GM ^{2} = 1 + ^{3}/_{4} = ^{7}/_{4}, so that
OM  = (Ö7)/2. Therefore,
sinÐOMC = sinÐOMG = OG /OM  = 2/(Ö7) . 

With P the midpoint of CD, the line MP (being the intersection
of the planes right bisecting AB and CD) passes through O.
Since MC and MD are altitudes of equilateral triangles,
MC  = MD  = 3(Ö3)/2, so that
MCD is isosceles with MP bisecting the apex angle. Hence
CD  = 2MC sin( 
1
2

ÐCMD) = 2 MC sinÐOMC = (6 Ö3)/(Ö7) . 

Solution 3. Let A be at (3/2, 0, 0) and
B be at (3/2, 0, 0) in space. The locus of points
equidistant from A and B is the plane x = 0. Let the
centre of the sphere be at the point (0, u, 0), so that
u^{2} + 9/4 = 4 and u = (Ö7)/2. Suppose that
C is at the point (0, y, z) where ^{9}/_{4} + y^{2} + z^{2} = 9. We have that (y  (Ö7)/2))^{2} + z^{2} = 4 whence
(Ö7)y + (9/4) = y^{2} + z^{2} = 9  (9/4) 

and y = 9/(2
Ö7). Therefore z = (3
Ö3)/(
Ö7).
We find that C ~ (0, 9/(2Ö7), (3Ö3)/(Ö7))
and D ~ (0, 9/(2Ö7), (3Ö3)/(Ö7)), whence
CD  = (6Ö3)/(Ö7).
Solution 4. [Y. Zhao] Use the notation of Solution 1, and note
that the right bisecting plane of AB and CD intersect in a
diameter of the sphere. Let us first determine the volume of the
tetrahedron ABDO. Consider the triangle ABD with centroid X.
We have that AB  = BD  = AD  = 3,
AX  = Ö3 and [ABD] = (9Ö3)/4. Since
O is equidistant from the vertices of triangle ABD, OX ^ABD. By Pythagoras' Theorem applied to triangle AXO, OX = 1, so that the volume of tetrahedron ABDO is
(1/3)OX [ABD] = (3Ö3)/4.
Triangle ABO has sides of lengths 2, 2, 3, and (by Heron's
formula) area (3Ö7)/4. Since CD ^ABO, CD is the
production of an altitude of tetrahedron ABDO; the altitude has
length ^{1}/_{2}CO . Hence
CD  = 
2[3Volume(ABDO)]
[ABO]

= 
2(9Ö3)/4
(3Ö7)/4

= 
6Ö3
Ö7

. 

Solution 5. [A. Wice] Let the tetrahedron ABCD have its
vertices on the surface of the sphere of equation x^{2} + y^{2} + z^{2} = 4 with A at (0, 0, 2). The remaining vertices have
coordinates (u, v, w) and satisfy the brace of equations:
u^{2} + v^{2} + (w  2)^{2} = 9 and u^{2} + v^{2} + w^{2} = 4. Hence,
w = 1/4. Thus, the points B, C, D lie on the circle of
equations z = 1/4, x^{2} + y^{2} = 63/16. The radius R of this
circle and the circumradius of triangle BCD is 3Ö7/4.
Triangle BCD has sides of length (b, c, d) = (b, 3, 3) and
area bcd/4R = ^{1}/_{2}bÖ{9  (b^{2}/4)}. Hence

3
Ö7

= 
1
2

 Ö

9  (b^{2}/4)

Û 36 = 7(9  (b^{2}/4)) Û144 = 7(36  b^{2}) Û b = 
6Ö3
Ö7

. 

Thus, the length of the remaining side is (6
Ö3)/
Ö7.

335.

Does the equation

1
a

+ 
1
b

+ 
1
c

+ 
1
abc

= 
12
a + b + c



have infinitely many solutions in positive integers a, b, c?
Comment. The equation is equivalent to
(a + b + c)(bc + ca + ab + 1) = 12abc . 

This is quadratic in each variable, and for any integer solution,
has integer coefficients. The general idea of the solution is to
start with a particular solution ((a, b, c) = (1, 1, 1) is an
obvious one), fix two of the variables at these values and regard the
equation as a quadratic in the third. Since the sum and the product
of the roots are integers and one root is known, one can find
another root and so bootstrap one's way to other solutions.
Thus, we have the quadratic for a:
(b + c)a^{2} + [(b + c)^{2} + (bc + 1)  12bc]a + (b + c)(bc + 1) = 0 , 

so that if (a, b, c) satisfies the equation, then so also does
(a
¢, b, c) where
a + a¢ = 
9bc  b^{2}  c^{2}  1
b + c

and aa¢ = bc + 1 . 

We can start constructing solutions using these relations:
(1, 1, 1), (1, 1, 2), (1, 2, 3), (2, 3, 7), (2, 5, 7),(3, 7, 11), (5, 7, 18), (5, 13, 18), ¼ . 

However, some triples do not lead to a second solution in integers.
For example, (b, c) = (2, 5) leades to
(7, 2, 5) and (11/7, 2, 5), and (b, c) = (3, 11) leads to
(7, 3, 11) and (34/7, 3, 11). So we have no guarantee that this
process will not peter out.
Solution 1. [Y. Zhao] Yes, there are infinitely many solutions.
Specialize to the case that c = a + b. Then the equation is
equivalent to
2(a + b)[(a + b)^{2} + ab + 1] = 12ab(a + b)Û a^{2}  3ab + b^{2} + 1 = 0 . 

This has at least one solution (a, b) = (1, 1). Suppose that
(a, b) = (p, q) is a solution with p
£ q. Then
(a, b) = (q, 3q
 p) is a solution. (To see this, note that
x
^{2}  3qx + (q
^{2} + 1) = 0 is a quadratic equation with one
root x = p and root sum 3q.) Observe that q < 3q
 p.
Define a sequence { x_{n} } for n ³ 0 by
x_{0} = x_{1} = 1 and x_{n} = 3x_{n1}  x_{n2} for n ³ 2.
Then (a, b) = (x_{0}, x_{1}) satisfies the equation, and by induction,
so does (a, b) = (x_{n}, x_{n+1}) for n ³ 1.
Since x_{n+1}  x_{n} = 2x_{n}  x_{n1}, one sees by induction that
{ x_{n} } is strictly increasing for n ³ 1. Hence, an
infinite set of solutions for the given equation is given by
(a, b, c) = (x_{n}, x_{n+1}, x_{n} + x_{n+1}) 

for n
³ 0. Some examples are
(a, b, c) = (1, 1, 2), (1, 2, 3), (2, 5, 7), (5, 13, 18), ¼ 

Comment. If { f_{n} } is the Fibonacci sequence defined by
f_{0} = 0, f_{1} = 1 and f_{n} = f_{n1} + f_{n2} for n ³ 2, then
the solutions are (a, b, c) = (f_{2k1}, f_{2k+1}, f_{2k1} +f_{2k+1}).
Solution 2. Yes. Let u_{0} = u_{1} = 1, v_{0} = 1, v_{1} = 2 and
u_{n} = 4u_{n1}  u_{n2} 

v_{n} = 4v_{n1}  v_{n2} 

for n
³ 2, so that { u
_{n} } = { 1, 1, 3, 11, 41,
¼}
and { v
_{n} } = { 1, 2, 7, 26, 97,
¼}. It can be proven by
induction that both sequences are strictly increasing for n
³ 1.
We prove that the equation of the problem is satisfied by
(a, b, c) = (u_{n}, v_{n}, u_{n+1}), (v_{n}, u_{n+1}, v_{n+1}) 

for n
³ 0. In other words, the equation holds if (a, b, c)
consists of three consecutive terms of the sequence
{ 1, 1, 1, 2, 3, 7, 11, 26, 41, 97,
¼}.
Observe that, for n ³ 2,

u_{n}v_{n}  u_{n+1}v_{n1} 

= u_{n}(4v_{n1}  v_{n2})  (4u_{n}  u_{n1})v_{n1} 
 
 
= u_{n1}v_{n1}  u_{n}v_{n2} 

so that, by induction, it can be established that
u
_{n} v
_{v}  u
_{n+1}v
_{n1} =
1. Similarly,

u_{n} v_{n+1}  u_{n+1}v_{n} 

= u_{n} (4v_{n}  v_{n1})  (4u_{n}  u_{n1})v_{n} 
 
 
= u_{n1} v_{n}  u_{n} v_{n1} = 1 . 

It can be checked that (a, b, c) = (1, 1, 1) satisfies the
equation. Suppose, as an induction hypothesis, that (a, b, c) = (u
_{n}, v
_{n}, u
_{n+1}) satisfies the equation. Then
u
_{n} is a root of the quadratic


= [x + v_{n} + u_{n+1}][x(v_{n} + u_{n+1}) +v_{n} u_{n+1} + 1]  12v_{n} u_{n+1} x 
 
 
= (v_{n} + u_{n+1})x^{2} + (v_{n}^{2} + u_{n+1}^{2} +1  9v_{n} u_{n+1})x + (v_{n} + u_{n+1})(v_{n}u_{n+1} + 1) . 

The second root is
v_{n+1} = 
v_{n} u_{n+1} + 1
u_{n}



so (a, b, c) = (v
_{n}, u
_{n+1}, v
_{n+1}) satisfies the equation.
Similarly, v
_{n} is a root of a quadratic, the product of whose
roots is u
_{n+1}v
_{n+1} + 1. The other root of this quadratic is
u_{n+2} = 
u_{n+1}v_{n+1} + 1
v_{n}



and so, (a, b, c) = (u
_{n+1}, v
_{n+1}, u
_{n+2}) satisfies the
equation.
Solution 3. [P. Shi] Yes. Let u_{0} = v_{0} = 1, v_{0} = 1,
v_{1} = 2, and
u_{n} = 
u_{n1}v_{n1} + 1
v_{n2}

and v_{n} = 
u_{n} v_{n1} + 1
u_{n1}



for n
³ 2. We establish that, for n
³ 1,
u_{n} = 2v_{n1}  u_{n1} ; 
 (1) 
v_{n} = 3u_{n}  v_{n1} ; 
 (2) 
3u_{n}^{2} + 2v_{n1}^{2} + 1 = 6u_{n}v_{n1} ; 
 (3) 
3u_{n}^{2} + 2v_{n}^{2} + 1 = 6u_{n}v_{n} . 
 (4) 
Note that (3u
_{n}^{2} + 2v
_{n}^{2} + 1
 6u
_{n}v
_{n})
(3u
_{n}^{2} + 2v
_{n1}^{2} + 1
 6u
_{n}v
_{n1}) = 2(v
_{n}  v
_{n1})(v
_{n} + v
_{n1}  3u
_{n}), so that the truth of any two of
(2), (3), (4) implies the truth of the third.
The proof is by induction. The result holds for n = 1.
Suppose it holds for 1 £ n £ k1.
From (3),
3u_{k1}^{2} + 2v_{k2}^{2} + 1 = 6u_{k1}v_{k2} Þ 

(3u_{k1}  v_{k2})u_{k1} + 1 = 2(3u_{k1}  v_{k2})v_{k2} u_{k1}v_{k2} . 

Substituting in (2) gives
u_{k} = 
u_{k1} v_{k1} + 1
v_{k2}

= 
u_{k1}(3u_{k1}  v_{k2}) + 1
v_{k2}

= 2v_{k1}  u_{k1} 

which establishes (1) for n = k. From (4),
(2v_{k1}  u_{k1})v_{k1} + 1 = [3(2v_{k1}  u_{k1})  v_{k1}]u_{k1} 

whence
v_{k} = 
u_{k} v_{k1} + 1
u_{k1}

= 
(2v_{k1}  u_{k1})v_{k1} + 1
u_{k1}

= 3u_{k} v_{k1} 

which establishes (2) for n = k.
From (4),
3(2v_{k1}  u_{k1})^{2} + 2v_{k1}^{2} + 1 = 6(2v_{k1}  u_{k1})v_{k1} 

Þ 3u_{k}^{2} + 2v_{k1}^{2} + 1 = 6u_{k}v_{k} .© 

Let w_{2n} = u_{n} and w_{2n+1} = v_{n} for n ³ 0.
Then
w_{n} = 
w_{n1}w_{n2} + 1
w_{n3}



for n
³ 3. It can be checked that (a, b, c) = (1, 1, 1) = (w
_{0}, w
_{1}, w
_{2}) satisfies the equation. Suppose that the
equation is satisfied by (a, b, c) = (w
_{n1}, w
_{n}, w
_{n+1}).
Thus
(x + w_{n} + w_{w+1})(x(w_{n} + w_{n+1}) + w_{n}w_{n+1} + 1) = 12xw_{n}w_{n+1} = 0 

is a quadratic equation in x one of whose roots is x = w
_{n1}.
The quadratic can be rewritten
x^{2} + [(w_{n} + w_{n+1})  (11w_{n}w_{n+1}  1)/(w_{n} + w_{n+1})]x+ (w_{n}w_{n+1} + 1) = 0 . 

Since the product of the roots is equal to w
_{n}w
_{n+1} + 1, the
second root is equal to

w_{n} w_{n+1} + 1
w_{n1}

= w_{n+2} . 

The result follows.
Solution 4. Let c = a + b. Then the equation becomes
2c(ab + c^{2} + 1) = 12abc, whence ab = (c^{2} + 1)/5. Hence,
a, b are roots of the quadratic equation
0 = t^{2}  ct + 
æ è


c^{2} + 1
5


ö ø

= 
1
4


é ë

(2t  c)^{2}  
æ è


5c^{2}  20
25


ö ø


ù û

. 

For this to have an integer solution, it is necessary that 5c
^{2} 20 = s
^{2}, the square of an integer s. Now s
^{2}  5c
^{2} =
20 is
a Pell's equation, three of whose solutions are (s, c) = (0, 2), (5, 3), (15, 7). Since x
^{2}  5y
^{2} = 1 is satisfied by
(x, y) = (9, 4), solutions (s
_{n}, c
_{n}) of s
^{2}  5c
^{2} =
20
are given by the recursion
s_{n+1} = 9s_{n} + 20c_{n} 

c_{n+1} = 4s_{n} + 9c_{n} 

for n
³ 0. where (s
_{0}, c
_{0}) is a starter solution.
Taking (s
_{0}, c
_{0}) = (0, 2), we get the solutions
(s; a, b, c) = (0; 1, 1, 2), (40; 5, 13, 18),(720; 89, 233, 322), ¼ . 

Taking (s
_{0}, c
_{0}) = (5, 3), we get the solutions
(s; a, b, c) = (5; 1, 2, 3), (105; 13, 34, 47),(1885; 233, 610, 843), ¼ . 

Taking (s
_{0}, c
_{0}) = (15, 7), we get the solutions
(s; a, b, c) = (15; 2, 5, 7), (275; 34, 89, 123), ¼ . 

Comment. Note that the values of c seem to differ from a
perfect square by 2; is this a general phenomenon?
Solution 5. [Z. Guo] Let r_{1} = r_{2} = 1, and, for n ³ 1,
r_{2n+1} = 2r_{2n}  r_{2n1} 

r_{2n+2} = 3r_{2n+1}  r_{2n} . 

Thus, { r
_{n} } = { 1. 1, 1, 2, 3, 7, 11, 26, 41, 87, 133,
¼}.
Observe that (a, b, c) = (r
_{m}, r
_{m+1}, r
_{m+2}) satisfies the
equation for m = 1, 2.
For each positive integer n, let k_{n} = r_{2n}  r_{2n1}, so that
r_{2n1} = r_{2n}  k_{n} , 

r_{2n+1} = r_{2n} + k_{n} , 

r_{2n+2} = 2r_{2n} + 3k_{n} . 

Suppose that (a, b, c) = (r_{2m1}, r_{2m}, r_{2m+1}) and
(a, b, c) = (r_{2m}, r_{2m+1}, r_{2m+2}) satisfy
the equation.
Substituting (a, b, c) = (r_{2m}  k_{m}, r_{2m}, r_{2m} + k_{m})
into the equation and simplifying yields that r_{2m} = Ö{3k_{m}^{2} + 1}. (This can be verified by substituting
(a, b, c) = (r_{2m}, r_{2m} + k_{m}, 2r_{2m} + 3k_{2m}) into
the equation.) In fact, this condition is equivalent to
these values of (a, b, c) satisfying the equation.
We have that


= 3(r_{2m+2}  r_{2m+1})^{2} + 1 
 
 
= 3(r_{2m} + 2k_{m})^{2} + 1 
 
 
= 3(r_{2m}^{2} + 4k_{m}^{2} + 4r_{2m}k_{m}) + 1 
 
 
= 3(7k_{m}^{2} + 1) + 12r_{2m}k_{m} + 1 = 21k_{m}^{2} + 12r_{rm} k_{m} + 4 . 

Thus


= (2r_{2m} + 3k_{m})^{2} = 4r_{2m}^{2} + 9k_{m}^{2} + 12r_{2m}k_{m} 
 
 
= 12k_{m}^{2} + 4 + 9k_{m}^{2} + 12r_{2m}k_{m} = 3k_{m+1}^{2} + 1 . 

This is the condition that
(a, b, c) = (r_{2m+2}  k_{m+1}, r_{2m+2}, r_{2m+2} + k_{m+1}) = (r_{2m+1}, r_{2m+2}, r_{2m+3}) 

and
(a, b, c) = (r_{2m+2}, r_{2m+2} + k_{m+1}, 2r_{2m+2} + 3k_{m+1}) = (r_{2m+2}, r_{2m+3}, r_{2m+4}) 

satisfy the equation. The result follows by induction.
Solution 6. [D. Rhee] Try for solutions of the form
(a, b, c) = 
æ è

x, 
1
2

(x + y), y 
ö ø



where x and y are postive integers of the same parity. Plugging this
into the equation and simplifying yields the equivalent equation
x^{2}  4xy + y^{2} + 2 = 0 Û x^{2}  4yx + (y^{2} + 2) = 0 . 

Suppose that z_{1} = 1, z_{2} = 3 and z_{n+1} = 4z_{n}  z_{n1}
for n ³ 1. Then, it can be shown by induction that z_{n+1} > z_{n} and z_{n} is odd for n ³ 1. We prove by induction that
(x, y) = (z_{n}, z_{n+1}) is a solution of the quadratic equation
in x and y.
This is true for n = 1. Suppose that it holds for n ³ 1. Then
z_{n} is a root of the quadratic equation x^{2}  4z_{n+1}x +(z_{n+1}^{2} + 2) = 0. Since the sum of the roots is the integer
4z_{n+1}, the second root is 4z_{n+1}  z_{n} = z_{n+2} and
the desired result holds because of the symmetry of the equation in
x and y.
Thus, we obtain solutions (a, b, c) = (1, 2, 3), (3, 7, 11),(11, 26, 41), (41, 97, 153), ¼ of the given equation.
Solution 7. [J. Park; A. Wice] As before, we note that
if (a, b, c) = (u, v, w) satisfies the equation, then so also
does (a, b, c) = (v, w, (vw + 1)/u). Define a sequence
{ x_{n} } by x_{1} = x_{2} = x_{3} = 1 and
x_{n+3} = 
x_{n+2}x_{n+1} + 1
x_{n}



for n
³ 1. We prove by induction, that for each n, the
following properties hold:
(a) x_{1}, ¼, x_{n+3} are integers; in
particular, x_{n} divides x_{n+2}x_{n+1} + 1;
(b) x_{n+1} divides x_{n} + x_{n+2};
(c) x_{n+2} divides x_{n} x_{n+1} + 1;
(d) x_{1} = x_{2} = x_{3} < x_{4} < x_{5} < ¼ < x_{n} < x_{n+1} < x_{n+2};
(e) (a, b, c) = (x_{n}, x_{n+1}, x_{n+2}) satisfies the equation.
These hold for n = 1. Suppose they hold for n = k. Since
x_{k+2}x_{k+3} + 1 = 
x_{k+2}(x_{k+1}x_{k+2} + 1)
x_{k}

+ 1 = 
x_{k+2}^{2} x_{k+1} + (x_{k} + x_{k+2})
x_{k}

, 

from (b) we find that x
_{k+1} divides the numerator. Since, by
(a), x
_{k} and x
_{k+1} are coprime, x
_{k+1} must divide
x
_{k+2}x
_{k+3} + 1.
Now
x_{k+1} + x_{k+3} = 
(x_{k} x_{k+1} + 1) + x_{k+1}x_{k+2}
x_{k}

. 

By (a), x
_{k} and x
_{k+2} are coprime, and, by (c), x
_{k+2}
divides the numerator. Hence x
_{k+2} divides x
_{k+1} +x
_{k+3}.
Since x_{k} = (x_{k+1}x_{k+2} + 1)/x_{k+3} is an integer,
x_{k+3} divides x_{k+1}x_{k+2} + 1. Next,
x_{k+3} = 
x_{k+1}x_{k+2} + 1
x_{k}

> 
æ è


x_{k+1}
x_{k}


ö ø

x_{k+2} > x_{k+2} . 

Finally, the theory of the quadratic delivers (e) for n = k + 1.
The result follows.

336.

Let ABCD be a parallelogram with centre O.
Points M and N are the respective midpoints of BO and
CD. Prove that the triangles ABC and AMN are similar
if and only if ABCD is a square.
Comment. Implicit in the problem, but what
should have been stated, is
that the similarity intended is in the order of the vertices as
given,
i.e., AB:BC:CA = AM:MN:NA. In the first solution,
other possible orderings of the vertices in the similarity are
considered (in which case, the result becomes false); in the
remaining solutions, the restricted sense of the similarity is
discussed.
Solution 1. Let P be the midpoint of CN. Observe that
[AMD] = 
3
4

[ABD] = 
3
4

[ABC] 

and
As N is the midpoint of CD, the area of triangle AMN is the
average of these two (why?), so that
Thus, in any case, we have the
ratio of the areas of the two triangles, so, if they are similar,
we know exactly what the factor of similarity must be.
Let AB  = CD  = 2a,
AD  = BC  = 2b, AC  = 2c and
BD  = 2d. Recall, that if UVW is a triangle with
sides 2u, 2v, 2w opposite the respective vertices U, V, W
and m is the length of the median from U, then
m^{2} = 2v^{2} + 2w^{2}  u^{2} . 

Since AN is a median of triangle ACD, with sides 2a, 2b, 2c,
AN ^{2} = 2c^{2} + 2b^{2}  a^{2} . 

Since AM is a median of triangle ABO, with sides 2a, c, d,
AM ^{2} = 2a^{2} + (c^{2}/2)  (d^{2}/4) . 

Since CM is a median of triangle CBO, with sides 2b, c, d,
CM ^{2} = 2b^{2} + (c^{2}/2)  (d^{2}/4) . 

Since MN is a median of triangle MCD with sides
CM
,
3d/2, 2a,


= 
CM ^{2}
2

+ 
1
2


æ è


3d
2


ö ø

2

 a^{2} 
 
 
= b^{2} + 
c^{2}
4

 
d^{2}
8

+ 
9d^{2}
8

 a^{2} = b^{2} + 
c^{2}
4

+ d^{2}  a^{2} . 

Suppose that triangle ABC and AMN are
similar (with any ordering of the
vertices). Then
8(AM ^{2} + AN ^{2} + MN ^{2}) = 5(AB ^{2} + AC ^{2} + BC ^{2}) 

Û 24b^{2} + 22c^{2} + 6d^{2} = 20a^{2} + 20b^{2} + 20c^{2} 

Û 2b^{2} + c^{2} + 3d^{2} = 10a^{2} . 
 (1) 
Also, for the parallelogram with sides 2a, 2b and diagonals
2c, 2d, we have that
c^{2} + d^{2} = 2(a^{2} + b^{2}) . 
 (2) 
Equations (1) and (2) together yield d
^{2} = 4a
^{2}  2b
^{2} and
c
^{2} = 4b
^{2}  2a
^{2}. We now have that

AN ^{2}
AC ^{2}

= 
2c^{2} + 2b^{2}  a^{2}
4c^{2}

= 
4c^{2} + c^{2}
8c^{2}

= 
5
8



which is the desired ratio. Thus, when triangles ABC and
AMN are similar, the sides AN and AC are in the correct
ratio, and so must correspond in the similarity. (There is a little
more to this than meets the eye. This is obvious when triangle
AMN is scalene; if triangle AMN were isosceles or equilateral,
then it is selfcongruent and the similarity can be set up to make
AN and BC correspond.)
We also have that

MN ^{2}
AB ^{2}

= 
4b^{2} + c^{2} + 4d^{2}  4a^{2}
16a^{2}

= 
4b^{2} + 4b^{2}  2a^{2} + 16a^{2}  8b^{2}  4a^{2}
16a^{2}

= 
10a^{2}
16a^{2}

= 
5
8



and

AM ^{2}
BC ^{2}

= 
8a^{2} + 2c^{2}  d^{2}
16b^{2}

= 
8a^{2} + 8b^{2}  4a^{2}  4a^{2} + 2b^{2}
16b^{2}

= 
10b^{2}
16b^{2}

= 
5
8

. 

Thus, if triangle ABC and AMN are similar, then
and cos
ÐABC = cos
ÐADC = 3(a
^{2}  b
^{2})/(2ab).
The condition that the cosine has absolute value not exceeding 1
yields that (
Ö{10}
 1)b
£ 3a
£ (
Ö{10} + 1)b.
Now we look at the converse. Suppose that ABCD is a parallelogram
with sides 2a and 2b as indicated above and that
cosÐABC = cosÐADC = 3(a^{2}  b^{2})/(2ab).
Then, the lengths 2c and 2d of the diagonals are given by
4c^{2} = AC ^{2} = 4a^{2} + 4b^{2}  12a^{2} + 12b^{2} = 16b^{2}  8a^{2} = 8(2b^{2}  a^{2}) 

and
4a^{2} = BD ^{2} = 4a^{2} + 4b^{2} + 12a^{2}  12b^{2} = 16a^{2}  8b^{2} = 8(2a^{2}  b^{2}) 

so that
AC
 = 2
Ö{4b
^{2}  2a
^{2}} and
BD
 = 2
Ö{4a
^{2}  2b
^{2}}.
Using the formula for the lengths of the medians, we have that
AN ^{2} = 2(4b^{2}  2a^{2}) + 2b^{2}  a^{2} = 5(2b^{2}  a^{2}) 

AM ^{2} = 2a^{2} + 
4b^{2}  2a^{2}
2

 
4a^{2}  2b^{2}
4

= 
5b^{2}
2



MN ^{2} = b^{2} + 
æ è


4b^{2}  2a^{2}
4


ö ø

+ (4a^{2}  2b^{2})  a^{2} = 
5a^{2}
2

. 

Thus

AN ^{2}
AC ^{2}

= 
AM ^{2}
BC ^{2}

= 
MN ^{2}
AB ^{2}

= 
5
8



and triangles ABC and AMN are similar with
AB:BC:AC=MN:AM:AN.
If triangles ABC and AMN are similar with AB:BC:AC=AM:MN:AN,
then we must have that AB = BC, i.e. a = b and so
ÐABC = ÐADC = 90^{°}, i.e., ABCD is
a square.
Comment. A direct geometric argument that triangles ABC and
AMN are similar when ABCD is a square can be executed as follows.
By reflection about an axis through M parallel to BC, we see that
MN = MC. By reflection about axis BD, we see that AM = CM
and ÐBAM = ÐBCM. Hence AM = MN and
ÐBAM = ÐBCM = ÐCMP = ÐNMP where P
is the midpoint of CN.
Consider a rotation with centre A through an angle BAM
followed by a dilation of factor AM /AB .
This transformation takes A® A, B ®M and the line BC to a line through M making an angle
ÐBAM with BC; the image line must be MN.
Since AB = BC and AM = MN, C ® N. Thus, the
image of triangle ABC is triangle AMN and the two triangles
are similar.
An alternative argument uses the fact that triangles AOM and ADN
are similar, so that ÐAMO = ÐAND and AMND is
concyclic. From this follows ÐAMN = 180^{°}  ÐADN = 90^{°}.
Solution 2. [S. Eastwood; Y. Zhao] If ABCD is a square,
we can take B at 1 and D at i, whereupon M is at
(3 + i)/4 and N at (1 + 2i)/2. The vector
[( ®)  MN] is represented by

1 + 2i
2

 
3 + i
4

= 
i(3 + i)
4

. 

Since [(
®)  AM] is represented by (3 + i)/4,
MN is obtained from AM by a 90
^{°} rotation about M so
that MN = AM and
ÐAMN = 90
^{°}. Hence the triangles
ABC and AMN are similar.
For the converse, let the parallelogram ABCD be represented
in the complex plane with A at 0, B at z and D at w,
Then C is at z + w, M is at (3z + w)/4 and N is at
^{1}/_{2}(z + w) + ^{1}/_{2}w = (z + 2w)/2.
Suppose that triangles ABC and AMN are similar. Then, since
ÐAMN = ÐABC and AM:AN = AB:AC, we must have that


= 
z
z + w

Û (3z + w)(z + w) = 2z(z + 2w) 

Û 3z^{2} + 4zw + w^{2} = 2z^{2} + 4zwÛ z^{2} + w^{2} = 0 Û z = ±iw . 

Thus AD is obtained from AB by a 90
^{°} rotation and
ABCD is a square.
Comment. Strictly speaking, the reasoning in the last
paragraph is reversible, so we could use it for the proof in
both directions. However, the particularization may aid in
understanding what is going on.
Solution 3. [F. Barekat] Let ABCD be a square. Then
DADC ~ DAOB so that AB:AC = OB:DC = MB:NC.
Since also ÐABM = ÐACN = 45^{°},
DAMB ~ DANC. Therefore AM:AB = AN:AC.
Also
ÐMAB = ÐNAC Þ ÐCAB = ÐCAM + ÐMAB = ÐCAM + ÐNAC = ÐNAM . 

Therefore
DAMN
~ DABC.
On the other hand, suppose that DAMN ~ DABC.
Then AM:AN = AB:AC and
ÐNAC = ÐNAM  ÐCAM = ÐCAB  ÐCAM = ÐMAB , 

whence
DAMB =
DANC.
Therefore, AB:AC = MB:NC = BO:DC. Since also ÐABO = ÐACD, DABO ~ DACD, so that
ÐABO = ÐACD and ÐAOB = ÐADC.
Because ÐABO = ÐACD, ABCD is a concyclic
quadrilateral and ÐDAB + ÐDCB = 180^{°}.
Since ABCD is a parallelogram, ÐDAB = ÐDCB = 90^{°} and ABCD is a rectangle. Thus ÐAOB = ÐADC = 90^{°}, from which it can be deduced that
ABCD is a square.
Solution 4. [B. Deng] Let ABCD be a square.
Since MN = MC = MA, triangles
MNC and AMN are isosceles. We have that
AM^{2} + MN^{2} = 2AM^{2} = 2(AO^{2} + OM^{2}) = (5/4)AB^{2} 

and
AN^{2} = AD^{2} + DN^{2} = (5/4)AB^{2} = AM^{2} + MN^{2} , 

whence
ÐAMN = 90
^{°} and
DAMN
~ DABC.
Suppose on the other hand that DAMN ~ DABC. Then
AN:AM = AC:AB and ÐNAM = ÐCAB together imply that
ÐNAC = ÐMAB Þ DNAC ~ DMABÞ ÐNCA = ÐABM . 

But
ÐNCA =
ÐOAB
Þ ÐOAB =
ÐOBA
Þ ABCD is a rectangle.
Let H be the foot of the perpendicular from M to CN. Then
ON, MH and BC are all parallel and M is the midpoint of
OB. Hence H is the midpoint of CN and DHMN º DHMC. Therefore MC = MN. Now, from the median length formula,
AM^{2} = 
1
4

(2BA^{2} + 2AO^{2}  BO^{2}) = 
1
4

(2BA^{2} + AO^{2}) 

and
CM^{2} = 
1
4

(2BC^{2} + 2CO^{2}  BO^{2}) = 
1
4

(2BC^{2} + CO^{2}) 

whence
(2BA^{2} + AO^{2}):(2BC^{2} + CO^{2}) = AM^{2}:CM^{2} = AM^{2}:MN^{2} = BA^{2}:BC^{2} 

so that
2BA^{2}·BC^{2} + AO^{2}·BC^{2} = 2BA^{2}·BC^{2} + BA^{2}·CO^{2}Þ BC^{2} = BA^{2} Þ BC = BA 

and ABCD is a square.

337.

Let a, b, c be three real numbers for which
0 £ c £ b £ a £ 1 and let w be a complex root of
the polynomial z^{3} + az^{2} + bz + c. Must w  £ 1?
Solution 1. [L. Fei] Let w = u + iv, [
`w] = u
 iv and r be the three roots. Then a =
2u
 r,
b =
w
^{2} + 2ur and c =
w
^{2} r.
Substituting for b, ac and c, we find that
w ^{6}  bw ^{4} + ac w ^{2}  c^{2} = 0 

so that
w
^{2} is a nonnegative real root of the
cubic polynomial q(t) = t
^{3}  bt
^{2} + act
 c
^{2} = (t
 b)t
^{2}+ c(at
 c). Suppose that t > 1, then t
 b and at
 c
are both positive, so that q(t) > 0. Hence
w
 £ 1.
Solution 2. [P. Shi; Y.Zhao]


= (1  w)(w^{3} + aw^{2} + bw + c) 
 
 
= w^{4} + (1  a)w^{3} + (a  b)w^{2} + (b  c)w + c 
 
 
Þ w^{4} = (1  a)w^{3} + (a  b)w^{2} + (b  c)w + c 
 
 
Þ w ^{4} £ (1  a)w ^{3}+ (a  b)w ^{2} + (b  c)w + c . 

Suppose, if possible, that
w
 > 1. Then
w ^{4} £ w ^{3}[(1  a) + (a  b) + (b  c) + c] = w ^{3} 

which implies that
w
 £ 1
and yields a contradiction. Hence
w
 £ 1.
Solution 3. There must be one real solution v. If v = 0, then
the remaining roots w and [`w], the complex conjugate of
w, must satisfy the quadratic equation z^{2} + az + b = 0.
Therefore w ^{2} = w[`w] = b £ 1 and the
result follows. Henceforth, let v ¹ 0.
Observe that
f(1) = 1 + a  b + c = (1  a)  (b  c) £ 0 

and that
f(c) = c^{3} + ac^{2}  bc + c ³ c^{3} + c^{3}  bc + c = c(1  b) ³ 0 , 

so that
1
£ v
£ c. The polynomial can be factored as
where c =
qv so that q = c/(
v)
£ 1. But q = w[
`w],
and the result again follows.

338.

A triangular triple (a, b, c) is a set of
three positive integers for which T(a) + T(b) = T(c). Determine
the smallest triangular number of the form a + b + c where
(a, b, c) is a triangular triple. (Optional investigations:
Are there infinitely many such triangular numbers a + b + c?
Is it possible for the three numbers of a triangular triple to
each be triangular?)
Solution 1. [F. Barekat]
For each nonnegative integer k, the triple
(a, b, c) = (3k + 2, 4k + 2, 5k + 3) 

satisfies T(a) + T(b) = T(c). Indeed,
(3k+2)(3k+3) + (4k+2)(4k+3) = (9k^{2}+15k+6)+(16k^{2}+20k+6) = 25k^{2} + 35k + 12 = (5k + 3)(5k + 4) . 

We have that a + b + c = 12k + 7, so we need to determine
whether there are triangular numbers congruent to 7 modulo 12.
Suppose that T(x) is such. Then x(x + 1) must be congruent to
14 modulo 24. Now, modulo 24,
x^{2} + x  14 º x^{2} + x  110 = (x  10)(x + 11) º (x  10)(x  13) 

so T(x) leaves a remainder 7 upon division by 12 if and only if
x = 10 + 24m or x = 13 + 24n for some nonnegative integers
m and n. This yields
and
for nonnegative integers m and n. The smallest triples according
to these formula are (a, b, c) = (14, 18, 23) and
(a, b, c) = (23, 30, 38), with the respective values of
a + b + c equal to 55 = T(10) and 91 = T(13).
However, it may be that there are
others that do not come under this set of formulae.
The equation a(a+1) + b(b+1) = c(c+1) is equivalent to
(2a + 1)^{2} + (2b + 1)^{2} = (2c + 1)^{2} + 1. It is straightforward to
check whether numbers of the form n^{2} + 1 with n odd is the sum
of two odd squares. We get the following triples with sums not
exceeding T(10) = 55:
(2, 2, 3), (3, 5, 6), (5, 6, 8), (4, 9, 10), (6, 9, 11), (8, 10, 13),(5, 14, 15), 

(9, 13, 16), (11, 14, 18), (6, 20, 21), (12, 17, 21), (9, 21, 23), (11, 20, 23), (14, 18, 23) . 

The entries of none except the last of these sum to a triangular
number.
Solution 2. [Y. Zhao] Let (a, b, c) be a triangular
triple and let n = a + b + c. Now


 2T(nab) = a^{2} + a + b^{2} + b  (n  a  b)^{2}  (n  a  b) 
 
 
= (n+1)(n+2)  2(n+1a)(n+1b) . 

Thus, (a, b, n
a
b) is a triangular triple if and only if
(n+1)(n+2) = 2(n+1
a)(n+1
b). In this case, neither n+1 nor
n+2 can be prime, as each factor on the right side is strictly less
than either of them. When 1 and 2 are added to each of the first
nine triangular numbers, 1, 3, 6, 10, 15, 21, 28, 36, 45, we get
at least one prime. Hence n
³ 55. It can be checked that
(a, b, c) = (14, 18, 23) is a triangular triple.
We claim that, for every nonnegative integer k, T(24k + 10) = a + b + c for some triangular triple (a, b, c). Observe that.
with n = T(24k + 10) = 288k^{2} + 252k + 55,


= (T(24k + 10) + 1)(T(24k + 10) + 2) 
 
 
= (288k^{2} + 252k + 56)(288k^{2} + 252k + 57) 
 
 
= 12(72k^{2} + 63k + 14)(96k^{2} + 84k + 19) . 

Select a and b so that
n+1a = 3(72k^{2} + 63k + 14) = 216k^{2} + 189k + 42 

and
n+1b = 2(96k^{2} + 84k + 19) = 192k^{2} + 168k + 38 . 

Let c = n
 a
 b. Thus,
(a, b, c) = (72k^{2} + 63k + 14, 96k^{2} + 84k + 18, 120k^{2} + 105k + 23) . 

From the first part of the solution, we see that (a, b, c) is a
triangular triple.
We observe that
(a, b, c) = (T(59), T(77), T(83)) = (1770, 3003, 3486) is a
triangular triple.
Check:


+ 3003 ×3004 = (2 ×3 ×7 ×11)(295 ×23 + 13 ×1502) 
 
 
= (2 ×3 ×7 ×11)(26311) = 2 ×3 ×7 ×11 ×83 ×317 
 
 
= (2 ×3 ×7 ×83)(11 ×317) = 3486 ×3487 . 

However, the sum of the numbers is 1770 + 3003 + 3486 = 8259,
which exceeds 8256 = T(128) by only 3.
Comment. David Rhee observed by drawing diagrams that
for any triangular triple (a, b, c), T(a + b  c) = (c  b)(c  a).
This can be verified directly. Checking increasing values of
a + b  c and factoring T(a + b  c) led to the smallest
triangular triple.