Solutions

227.

Let n be an integer exceeding 2 and let
a_{0}, a_{1}, a_{2}, ¼, a_{n}, a_{n+1} be positive real numbers for which
a_{0} = a_{n}, a_{1} = a_{n+1} and
a_{i1} + a_{i+1} = k_{i} a_{i} 

for some positive integers k_{i}, where 1 £ i £ n.


Prove that
2n £ k_{1} + k_{2} + ¼+ k_{n} £ 3n . 

Solution. Since k
_{i} = (a
_{i1}/a
_{i}) + (a
_{i+1}/a
_{i})
for each i,

n å
i=1

k_{i} = 
n å
i=1


æ è


a_{i+1}
a_{i}

+ 
a_{i}
a_{i+1}


ö ø

³ 
n å
i=1

2 = 2n . 

As for the other inequality, since the expression has cyclic
symmetry, there is no loss in generality in supposing that
a_{n} ³ a_{1} and a_{n} ³ a_{n1}
with inequality in at least one case, so that
2a_{n} > a_{n1} + a_{1}. Therefore, k_{n} = 1 and
a_{n} = a_{n1} + a_{1}.
We establish the right inequality by induction. For the
case n = 3, we may suppose that
a_{2} + a_{3} = k_{1} a_{1} ; a_{1} + a_{3} = k_{2} a_{2} ; a_{1} + a_{2} = a_{3} 

Substituting for a
_{3} and rearranging the terms yields
the brace of equations
2a_{2} = (k_{1}  1)a_{1} 2a_{1} = (k_{2}  1)a_{2} 

whence 4 = (k
_{1}  1)(k
_{2}  1). It follows that
k
_{1} + k
_{2} + k
_{3} is either 5 + 2 + 1 = 8 or 3 + 3 + 1 = 7.
Now suppose the result holds when the index is n  1 ³ 3.
Then, supposing that k_{n} = 1 and substituting for a_{n},
we obtain the n1 equations
a_{n1} + a_{2} = (k_{1}  1)a_{1} 

a_{1} + a_{3} = k_{2} a_{2} 

a_{n3} + a_{n1} = k_{n2}a_{n2} 

a_{n2} + a_{1} = (k_{n1}  1)a_{n1} . 

By the induction hypothesis
(k_{1}  1) + k_{2} + ¼+ (k_{n1}  1) £ 3(n  1) = 3n  3 

whence
k_{1} + k_{2} + ¼+ k_{n} £ (3n  3) + 2 + 1 = 3n . 



241.

Determine sec40^{°} + sec80^{°}+ sec160^{°}.
Solution 1. The values 40^{°}, 80^{°} and
160^{°} all satisfy cos3q = 3D 1/2, or
8 cos^{3} q 6cosq+ 1 = 3D 0. Thus,
cos40^{°}. cos80^{°} and cos160^{°}
are the roots of the cubic equation 8x^{3}  6x + 1 = 3D 0,
so that their reciprocals sec40^{°},
sec80^{°} and sec160^{°} are the roots of
the cubic equation x^{3}  6x^{2} + 8 = 3D 0. The sum of the roots
of this cubic is
sec40^{°} + sec80^{°} + sec160^{°} = 3D 6 . 

Solution 2. Let z = 3D cos40^{°} + i sin 40^{°}. Then z^{9} = 3D 1. In fact, since z^{9}  1 = 3D (z  1)(z^{2} + z + 1) (z^{6} + z^{3} + 1) and the first two factors
fail to vanish, z^{6} + z^{3} + 1 = 3D 0. Also 1 + z + z^{2} + ¼+ z^{8} = 3D (1 + z + z^{2})(1 + z^{3} + z^{6}) = 3D 0. Observe that
cos40^{°} = 3D ^{1}/_{2}(z + ^{1}/_{z}),
cos80^{°} = 3D ^{1}/_{2}(z^{2} + [ 1/(z^{2})])
and cos160^{°} = 3D ^{1}/_{2}(z^{4} + [ 1/(z^{4})]),
so that the given sum is equal to

2 
é ë


z
1 + z^{2}

+ 
z^{2}
1 + z^{4}

+ 
z^{4}
1 + z^{8}


ù û



=3D 2 
é ë


z
1 + z^{2}

+ 
z^{2}
1 + z^{4}

+ 
z^{5}
1 + z


ù û


 
 
=3D 2 
é ë


z(1 + z + z^{4} + z^{5}) + z^{2}(1 + z + z^{2} + z^{3}) + z^{5}(1 + z^{2} + z^{4} + z^{6})
(1 + z)(1 + z^{2})(1 + z^{4})


ù û


 
 
=3D 2 
é ë


z^{7} + z^{6} + 3z^{5} + z^{4} + z^{3} + 3z^{2} + z + 1
(1 + z)(1 + z^{2})(1 + z^{4})


ù û


 
 
=3D 2 
é ë


(z+1)(z^{6} + z^{3} + 1) + 3z^{2} (z^{3} + 1)
(1 + z)(1 + z^{2})(1 + z^{4})


ù û


 
 
=3D 2 
é ë


0  3z^{8}
1 + z + z^{2} + z^{3} + z^{4} + z^{5} + z^{6} + z^{7}


ù û

=3D 2 
é ë


3z^{8}
z^{8}


ù û

=3D 6 . 

Solution 3. [T. Liu]

sec40^{°} + sec80^{°} + sec 160^{°} 

=3D 
cos40^{°} + cos80^{°}
cos40^{°} cos80^{°}

+ 
1
cos 160^{°}


 
 
=3D 
2cos60^{°} cos20^{°}
cos40^{°}cos80^{°}

+ 
1
cos160^{°}


 
 
=3D 
cos20^{°} cos160^{°} + cos40^{°} cos80^{°}
cos40^{°} cos80^{°} cos 160^{°}


 
 
=3D 
cos180^{°} + cos140^{°} + cos120^{°} + cos40^{°}
cos40^{°}(cos240^{°} + cos80^{°})


 
 
=3D 
1  1/2
(1/2)(cos40^{°} + cos120^{°}+ cos40^{°})

=3D 
3/2
1/4

=3D 6 . 

Solution 4. Let x = 3D cos40^{°}, y = 3D cos 80^{°} and z = 3D cos160^{°}. Then
x + y + z = 3D 2cos60^{°}cos20^{°}  cos20^{°} = 3D 0 

and


=3D 
1
2

[cos120^{°} + cos140^{°} + cos240^{°} + cos 80^{°} + cos200^{°} + cos120^{°}] 
 
 
=3D 
1
2


é ë

 
3
2

+ x + y + z 
ù û

=3D  
3
4

. 

Now

8 sin40^{°} cos40^{°} cos80^{°} cos160^{°} 

=3D 4 sin80^{°} cos80^{°} cos 160^{°} 
 
 
=3D 2 sin160^{°} cos160^{°} = 3D sin 320^{°} = 3D sin40^{°} 

so that xyz = 3D 1/8. Then the sum of the problem is equal to
(xy + yz + zx)/(xyz) = 3D 6.
248.

Find all real solutions to the equation
Solution 1. For the equation to be valid over the reals,
we require that x
³ 1. Suppose that y
^{2} = x
 1 and
y
³ 0. Then the equation becomes
When 1
£ x
£ 5, we have that 0
£ y
£ 2
and the equation becomes (2
 y) + (3
 y) = 1 or y = 2, x = 5.
When 5
£ x
£ 10, we have that 2
£ y
£ 3 and
the equation becomes an identity (y
 2) + (3
 y) = 1.
Thus, it holds for all x on the closed interval [5, 10].
Finally, when 10
£ x, we have that 3
£ y and the
equation becomes (y
 2) + (y
 3) = 1 or y = 3, x = 10.
Thus, the complete set of solutions of the equation is given
by 5
£ x
£ 10. All these solutions check out.
Solution 2. [Z. Wu] For a solution to exist, we require
that x ³ 1 and that both 0 £ x + 3  4Ö{x  1} £ 1 and 0 £ x + 8  6Ö{x  1} £ 1. These two
conditions lead to (x + 2)^{2} £ 16(x  1) and
(x + 7)^{2} £ 36(x  1), which in turn leads to
(x  2)(x  10) = (x^{2} + 4x + 4)  (16x  16) £ 0 

and
(x  5)(x  17) = (x^{2} + 14x + 49)  (36x  36) £ 0 . 

These conditions are both satisfied only if 5
£ x
£ 10.
(Thus, 5
£ x
£ 10 is a
necessary condition for
a solution.)
On the other hand, 5 £ x £ 10 implies that 2 £ Ö{x1} £ 3, so that (as in Solution 1) we find that the equation is
equivalent to (Ö{x  1}  2) + (3  Ö{x1}) = 1, which
is an identity. Thus, the equation holds exactly when
5 £ x £ 10.
Comment. Your first observation should be that, in order
for the equation to make sense, we require that x ³ 1. It
is important not just to write down a lot of algebraic equations,
but to indicate the logical relationships between them; which
equations imply which other equations? which pairs of equations
are equivalent? This is especially desirable when surd equations
are involved, where the operations that lead from one equation
to another are not logically reversible and extraneous solutions
might be introduced.

249.

The nonisosceles right triangle ABC has
ÐCAB = 90^{°}. Its inscribed circle with centre
T touches the sides AB and AC at U and V respectively.
The tangent through A of the circumscribed circle of
triangle ABC meets UV
in S. Prove that:
(a) ST
 BC;


(b) d_{1}  d_{2}  = r , where
r is the radius of the inscribed circle, and d_{1} and d_{2} are the
respective distances from S to AC and AB.
Solution. Wolog, we may assume that AB < AC so that
S and C are on opposite sides of AB.
Ad (a),
ÐSVT =
ÐSVA = 45
^{°}, AV = VT and
SV is common, so that triangles AVS and TVS are congruent.
Hence
ÐSAV =
ÐSTV
ÞÐSTU =
ÐSAU =
ÐACB (by the tangentchord
property). Since TU
 AC, it follows that CB
 ST.
Ad (b), let P and Q be the respective feet of the
perpendiculars from S to AB and AC.
Note that SQAP is a rectangle so that
ÐPUS = ÐPSU = 45^{°} and so PU = PS.
Then QS  PS  = AP  PU  = r.

250.

In a convex polygon P, some diagonals
have been drawn so that no two have an intersection in the interior
of P. Show that there exists at least two vertices of
P, neither of which is an enpoint of any of these diagonals.
Solution 1.
If no diagonal has been drawn, the result is clear.
Suppose that at least one diagonal has been drawn. Let d be a
diagonal that has, on one of its sides, the fewest vertices of
the polygon. There is at least one such vertex.
Then on that side, no further diagonal is drawn,
since it cannot cross d and cannot have fewer vertices between
its endpoints than d. Hence there is at least one vertex
on that side from which no diagonal is drawn.
On the other side of d, select a diagonal g which has the
smallest number of vertices between its endpoints on the side
opposite to the side of d. By an argument similar to the above,
there is at least one vertex on the side of g opposite to
d from which no diagonal has been drawn.
Solution 2. [S. King] The result is vacuously true for
triangles. Suppose that the polygon has at least four sides.
Suppose that a (possibly void) collection of diagonals as
specified in the problem is given. We continue adding diagonals
one at a time such that each new diagonal does not cross any
previous one in the interior of the polygon. At each stage,
the polygon P is partitioned into polygons with
fewer sides all of whose vertices are vertices of the
polygon P. As long as any of the subpolygons has more
than three sides, we can add a new diagonal. However, the process
will eventually terminate with a triangulation of P,
i.e., a partitioning of P into n  2 triangles
all of whose vertices are vertices of P. (Exercise.
Explain why the number of triangles is n  2. One way to do
this is to note that the sum of all the angles of the triangles
is equal to the sum of the angles in the polygon.)
Each triangle must have at most two sides in common with the
given polygon. Since there are n sides and n  2 triangles,
at least two triangles have two sides in common with P.
In each case, the vertex common to the two sides has no diagonal
emanating from it (neither an original diagonal nor an added
diagonal), and the result follows.
Comment. Many solvers failed to appreciate that the collection
of diagonals is given, and that the problem is to establish the
desired property no matter what the collection is. A lot of
arguments had the students constructing diagonals without indicating
how the ones constructed might have anything to do with a given set;
in effect, they were giving a particular situation in which the
result obtained. Several solvers tried induction, using one diagonal
to split P into two, but did not handle well the possibility
that the loose vertices in the subpolygons might be at the ends of
the subdividing diagonal. One way around this is to make the
result stronger, and show that one can find two nonadjacent
vertices that are not the endpoints of diagonals. This is certainly
true for quadrilaterals, and using this an induction hupothesis yielded
a straightforward argument for polygons of higher order.

251.

Prove that there are infinitely many positive integers
n for which the numbers { 1, 2, 3, ¼, 3n } can be
arranged in a rectangular array with three rows and n columns for which
(a) each row has the same sum, a multiple of 6, and (b) each
column has the same sum, a multiple of 6.
Solution 1. The sum of all the numbers in the array is
3n(3n+1)/2, so that each column sum must be 3(3n + 1)/2.
Since this is divisible by 6, 3(3n+1) must be a multiple of
12, and so 3n + 1 is divisible by 4 and n
º 1 (mod 4).
Since each row sum, n(3n+1)/2 is divisible by 6, n must
be divisible by 3. Putting this together, we conclude that
n = 12k + 9 for some value of k.
We now show that, for each n of this form, we can actually
construct an array with the desired property. Starting with
the magic square, we derive the following array for n = 9:
We generalize this for n = 12k + 9, for k a nonnegative
integer. Suppose that an array is possible. Then the sum
of all the elements in the array is (36k + 27)(18k + 14) = 18(4k + 3)(9k + 7). The sum of the elements in each column
is 6(9k + 7) = 54k + 42 and the sum in each row is
6(4k + 3)(9k + 7) = (4k + 3)(54k + 42). If we can achieve
this with distinct entries, then we have constructed the array.
We build the array by juxtaposing horizontally
4k + 3 square 3 ×3 blocks of the form:
where we make 4k + 3 distinct choices of each of
a, b, c to ensure that no number is repeated in any
row (it is not possible for any repetition to occur
down a column). To achieve the column sum, we
require that 15 + 9(a + b + c) = 54k + 42, or
a + b + c = 6k + 3 = 3(2k + 1). To achieve the row
sum, we require that


= 15(4k + 3) + 27 
å
 b = 15(4k + 3) + 27 
å
 c 
 
 

so that

å
 a = 
å
 b = 
å
 c = (4k + 3)(2k + 1) = 0 + 1 + ¼+ (4k + 2) , 

where each sum is over 4k + 3 distinct elements. It is
convenient to let the sets of a's, b's, and c's
each consist of the numbers 0, 1, 2,
¼, 4k + 2 in
some order. In the ith 3 ×3 block, let
0
£ a, b, c,
£ 4k + 2 and
b º (i  1) + 2(k + 1) = 2k + i + 1 (mod 4k + 2) 

c = (6k + 3)  (a + b) º 4k + 3  2i (mod 4k + 3) . 

for 1
£ i
£ 4k+3. It is straightforward to verify that
the a's, the b's and the c's each
run through a complete set of residues (mod 4k + 3),
and we have arranged that a + b + c = 6k + 3. If
1
£ i
£ 2k + 1, then 2k + 2
£ b = 2k + i + 1
£ 4k + 2
and 2k + 2
£ a + b = 2(k + i)
£ 6k + 2, so that 1
£ c
£ 4k + 1. If 2k + 2
£ i
£ 4k + 3, then
0
£ b = (2k + i + 1)
 (4k + 3) = i
 2k
 2
£ 2k + 1
and 2k + 1
£ a + b = 2i
 2k
 3
£ 6k + 3, so that
0
£ c
£ 4k + 2. With this choice of the variables
a, b, c we can construct the array as desired.
For example, when n = 45, k = 3, there are 15 blocks and the
choice of a, b, c for these blocks can be read along the rows
of
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
8 
9 
10 
11 
12 
13 
14 
0 
1 
2 
3 
4 
5 
6 
7 
13 
11 
9 
7 
5 
3 
1 
14 
12 
10 
8 
6 
4 
2 
0 
It is left as an exercise for the reader to construct the 3 ×45
array.
Solution 2. [Y. Zhao] We can form the 3 ×9 array:
Suppose, as an induction hypothesis, we can build a 3 ×n array
for some positive integer n. Duplicate this array five times and
put them side by side in a row. Partition the 3 ×5n array
into fifteen 1 ×n subarrays, and to the elements of
each of the fifteen subarrays add a constant number as indicated
by the positions in the following 3 ×5 table:
The row sum of the numbers added is 30n and the column sum is
18n, so the 3 ×5n array preserves the divisibility by
6 properties of the 3 ×n array. Therefore, we can
see by induction that an array is constructibel whenever
n = 9 ×5
^{k} for 0
£ k.
Solution 3. [J. Zhao] For the time being, neglect the conditions
involving divisibility by 6, and focus only on the condition that
the numbers 1, 2, ¼, 3n be used and that the row sums and
the column sums be each the same. Then, when n = 3, a magic
square will serve.
Suppose that, for some k ³ 1, we have found a suitable
3 ×3^{k} matrix M. Let A be the 3 ×3^{k+1}
matrix obtained by placing three copies of M side by side and
B the 3 ×3^{k+1} matrix determined by placing side
by side the 3 ×3^{k} matrices B_{1}, B_{2}, B_{3} where
each column of B_{1} is (the transpose of) (0, 1, 2), of B_{2}
is (1, 2, 0), and of B_{3} is (2, 0, 1). Each of A and
B has constant row sums and constant column sums.
Let N = A + 3^{k+1}B. Then N not only has constant row
and column sums, but consists of the numbers 1, 2, ¼, 3^{k+2}
(why?). The row sums of M are each (1/6)(3^{k+1})(3^{k+1} + 1),
so that the row sums of N are each

3 ×(1/6)(3^{k+1})(3^{k+1} + 1) 

+ 3^{k+1}(3^{k}) + 3^{k+1}(2 ×3^{k}) = (1/6)[3^{k+2}(3^{k+1} + 1)] + (3^{2k+1} ×3) 
 
 
= (1/6)(3^{k+2})(3^{k+1} + 1 + 6 ×3^{k}) = (1/6)(3^{k+2})(3^{k+2} + 1) . 

The column sums of M are each (3/2)(3
^{k+1} + 1) and so the
column sums of N are each
(3/2)(3^{k+1} + 1) + 3^{k+1} + 2 ×3^{k+1} = (1/2)(3^{k+2} + 3 + 2 ×3^{k+2}) = (3/2)(3^{k+2} + 1) . 

We now require that each of (1/6)(3^{k+1})(3^{k+1} + 1) and
(3/2)(3^{k+1} + 1) be divisible by 6. This will occur exactly
when 3^{k+1} + 1 º 0 (mod 4), so that k must be even.
Thus, we can obtain an array as desired when n = 9^{m} for
some positive integer m. (Note that 9^{m} º 9 (mod 12).)

252.

Suppose that a and b are the roots of the
quadratic x^{2} + px + 1 and that c and d are the roots of
the quadratic x^{2} + qx + 1. Determine
(a  c)(b  c)(a + d)(b + d) as a function of p and q.
Solution 1. From the theory of the quadratic, we have that
a + b =
p, c + d =
q and ab = cd = 1. Then


(a + d)(b + d) = (a  c)(b + d)(b  c)(a + d) 
 
 
= (ab  cd + ad  bc)(ba  cd + bd  ca) 
 
 
= (ad  bc)(bd  ca) = abd^{2}  a^{2}cd  b^{2}cd + abc^{2} 
 
 
= d^{2}  a^{2}  b^{2} + c^{2} = [(c + d)^{2}  2cd]  [(a + b)^{2}  2ab] 
 
 
= (q^{2}  2)  (p^{2}  2) = q^{2}  p^{2} . 

Solution 2. Using a + b = p, c + d = q and ab = cd = 1,
we obtain that


= [ab  (a + b)c + c^{2}][ab + (a + b)d + d^{2}] 
 
 
= (1 + pc + c^{2})(1  pd + d^{2}) = (2 + c^{2} + d^{2})  p^{2} 
 
 
= (c + d)^{2}  p^{2} = q^{2}  p^{2} . 


253.

Let n be a positive integer and let q = p/(2n+1). Prove that cot^{2} q, cot^{2} 2q,
¼, cot^{2} nq are the solutions of the equation

æ è

2n + 1
1

ö ø

x^{n}  
æ è

2n + 1
3

ö ø

x^{n1} + 
æ è

2n+1
5

ö ø

x^{n2} ¼ = 0 . 

Solution 1. From de Moivre's Theorem that
cosmq+ i sinmq = (cosq+ i sinq)^{m} , 

we obtain from a comparison of imaginary parts that
sinmq = 
æ è

m
1

ö ø

cos^{m1}qsinq 
æ è

m
3

ö ø

cos^{m3} sin^{3} q+ ¼ , 

for each positive integer m.
Hence
sin(2n + 1)q = sin^{2n+1}q 
é ë


æ è

2n + 1
1

ö ø

cot^{2n} q 
æ è

2n + 1
3

ö ø

cot^{2n2} q+ ¼ 
ù û

. 

When
q = (k
p)/(2n + 1) for 1
£ k
£ n,
sin(2n + 1)
q = 0 while sin
q ¹ 0.
The desired result follows.
Solution 2. [Y. Zhao] Observe that, for each complex a,

1
2

[(a + 1)^{2n+1}  (a  1)^{2n+1}] = 
æ è

2n + 1
1

ö ø

a^{2n} + 
æ è

2n + 1
3

ö ø

a^{2n  2} + 
æ è

2n + 1
5

ö ø

a^{2n  4} + ¼ . 

Suppose that a = i cotk
q with
q =
p/(2n+1) and
1
£ k
£ n. Note that sink
q ¹ 0. Then


æ è

2n + 1
1 (cot^{2} kq)^{n} 

+ 
æ è

2n + 1
3 (cot^{2} kq)^{n1} + ¼ = 
1
2

[ (i cotk q+ 1)^{2n + 1} (i cotq 1)^{2n + 1}] 
 
 
= 
1
2


æ è


i
sinkq


ö ø

2n + 1

[ (coskq i sinkq)^{2n + 1} (coskq+ i sinkq)^{2n + 1} ] 
 
 
= 
æ è


i
sinkq


ö ø

2n + 1

[  sin(2n + 1)k q] = 
æ è


i
sinkq


ö ø

2n + 1

[  sinkp] = 0 , 

and the result follows.

254.

Determine the set of all triples (x, y, z) of integers
with 1 £ x, y, z £ 1000 for which x^{2} + y^{2} + z^{2} is a
multiple of xyz.
Solution. Suppose that x
^{2} + y
^{2} + z
^{2} = kxyz, for
a positive integer k. It
can be checked that if the equation is satisfied by
(x, y, z) then it is also satisfied by (x, y, kxy
 z).
Since z
^{2} < x
^{2} + y
^{2} + z
^{2} = kxyz, it follows that
z < kxy, If z >
^{1}/
_{2}kxy, then kxy
 z <
^{1}/
_{2}kxy. Suppose that we start with a solution.
If we have, say z exceeding
^{1}/
_{2}xy, then we can
replace z by a new value less than
^{1}/
_{2}xy.
We can do the analogous thing with x and y. Every such
operation reduces the sum x + y + z, so it can be performed
at most finitely often, and we reach a situation where it
cannot be done any more. Thus, we arrive at a solution
where, say,
1
£ x
£ y
£ z
£ kxy/2, so that, in particular
kx
³ 2. We can also start with such a solution and
go backwards to achieve any given solution.
Since
x^{2} + y^{2} + 
æ è


kxy
2

 z 
ö ø

2

= 
æ è


kxy
2


ö ø

2

, 

it follows that
x^{2} + y^{2} + 
æ è


kxy
2

 y 
ö ø

2

³ 
æ è


kxy
2


ö ø

2

, 

so that
3y^{2} ³ x^{2} + 2y^{2} ³ kxy^{2} 

and kx
£ 3. Thus kx = 2 or kx = 3.
The case kx = 2 leads to x^{2} + (y  z)^{2} = 0 which has
no solutions as specified. Hence kx = 3 and k = 1 or k = 3.
For these two cases, we find that the base solutions are
respectively (x, y, z) = (3, 3, 3) and (x, y, z) = (1, 1, 1).
Suppose that k = 1. Modulo 3, any square is congruent to 0
or 1. If, say, x º 0 (mod 3), then y^{2} + z^{2} º 0
(mod 3); this can occur only if y and z are multiples of
3. Hence (x, y, z) = (3u, 3v, 3v) for some integers
u, v, w. But then 9u^{2} + 9v^{2} + 9w^{2} = 27uvw, or
u^{2} + v^{2} + w^{2} = 3uvw. Contrarily, any solution (u, v, w)
of this equation gives rise to a solution (x, y, z) of
x^{2} + y^{2} + z^{2} = xyz. Therefore there is a oneone
correspondence between solutions of x^{2} + y^{2} + z^{2} = xyz
where all numbers are multiples of 3 and solutions of
x^{2} + y^{2} + z^{2} = 3xyz. We will obtain these solutions below.
The only other possibility is that none of x, y, z is
divisible by 3. But then, x^{2} + y^{2} + z^{2} would be a multiple
of 3 and xyz not a multiple of 3; thus there are no solutions
of this type.
Suppose that k = 3. From the above, we know that every solution
arises from a solution for which 1 £ x £ y £ z £ 3xy/2
and for such a solution x = 1. Let (x, y, z) = (1, y, ty) where 1 £ y £ 3/2. Then
1 + (1 + t^{2})y^{2} = 3ty^{2}, so that
y^{2} = 
1
3t  1  t^{2}

= 
1

. 

The denominator is not less than 1, so that y
^{2} £ 1.
Hence the only solution that can generate the rest is
(x, y, z) = (1, 1, 1).
To get a handle on the situation, fix x = u and consider
a sequence of solutions (x, y, z) = (u, v_{n1}, v_{n}).
The solution w = v_{n1} satisfies the quadratic equation
u^{2} + w^{2} + v_{n}^{2} = 3uwv_{n} 

and so also does a second value w = v
_{n+1}. By the
theory of the quadratic, we have that
v_{n+1} + v_{n1} = 3uv_{n} 
 (1) 
and
v_{n+1} v_{n1} = u^{2} + v_{n}^{2} . 
 (2) 
If we start off with a solution (x, y, z) = (u, v
_{0}, v
_{1}),
we can use either (1) or (2) to determine the sequence
{ v
_{n} }. Note that, since
v_{n+1}  v_{n} = v_{n}  v_{n1} + (3u  2)v_{n} > v_{n}  v_{n1} , 

if v
_{1} ³ v
_{0}, then the sequence { v
_{n} } is increasing.
Note also, that the equations (1) and (2) are symmetric
in v
_{n1} and v
_{n+1}, so
we can extend the sequence backwards as well as forwards.
Using the recursion v_{n+1} = 3uv_{n}  v_{n1}, we get the
following sequences for various values of u:
u = 1 : { v_{n} } = { 1, 1, 2, 5, 13, 34, 89, 233, 610 } 

u = 2 ; { v_{n} } = { 1, 1, 5, 29, 169, 985 } 

u = 5 ; { v_{n} } = { 194, 13, 1, 2, 29, 433 } 

u = 13 ; { v_{n} } = { 34, 1, 5, 194 } 

u = 29 ; { v_{n} } = { 433, 5, 2, 169 } 

u = 34 ; { v_{n} } = { 13, 1 89 } 

and so on. This yields the following solutions with
1
£ x
£ y
£ z
£ 1000:
(x, y, z) = (1, 1, 1), (1, 1, 2), (1, 5, 13),
(1, 13, 34), (1, 34, 89), (1, 89, 233),
(1, 233, 610), (2, 5, 29), (2, 29, 169),
(2, 169, 985), (5, 13, 194), (5, 29, 433).