Page 1 of 5

20 At Right Angles | Vol. 3, No. 1, March 2014

feature

Harmonic Triples

Part 3

Observe a relationship, then prove it – satisfying in itself.

Take this one step further and find the geometrical connect.

Excitement squared!

Shailesh Shirali

PHTs . . . Primitive and beautiful

I

n Parts–1, 2 of this article which appeared in the March 2013 and July

2013 issues of At Right Angles, we introduced the notion of a primitive

harmonic triple (‘PHT’) as a triple (a, b, c) of coprime positive integers

satisfying the equation

Examples: (3,6,2) and (6,30,5). We showed that the equation surfaces

in many contexts, and we explored ways of generating PHTs. Now we

explore some properties of these triples. Table 1 lists many of the triples.

(To avoid duplication we have added the condition a ≤ b.) It is worth

studying the list to identify features of interest.

(2,2,1), (3,6,2), (4,12,3), (5,20,4),

(6,30,5), (7,42,6), (8,56,7), (9,72,8),

(10,15,6), (10,90,9), (14,35,10), (18,63,14),

(21,28,12), (22,99,18), (24,40,15), (30,70,21),

(33,88,24), (36,45,20), (44,77,28), (55,66,30),

(60,84,35), (65,104,40), (78,91,42), (105,120,56).

Table 1. A list of some primitive harmonic triples (PHTs)

Keywords: Primitive harmonic triples, perfect squares, rhombus, triangle

Page 2 of 5

At Right Angles | Vol. 3, No. 1, March 2014 21

Perfect squares

Among the many noticeable features of PHTs, the one that strikes the eye most is the presence of numerous

perfect squares associated with each triple.

Proposition 1 If (aaaaaa aaaaaa aaaaaa is a primitive harmonic triple, then aaaa aa aaaa, aaaa aa aaaa and aaaa aa aaaa are perfect squares.

Example: Take the PHT (10aa 15aa 6aa; we have: 10 aa 15 = 5�, 10 aa 6 = 2�, 15 aa 6 = 3�. But still more

can be said.

Proposition 2 If (aaaaaa aaaaaa aaaaaa is a primitive harmonic triple, then aaaaaaaaaaaa is a perfect square.

Example: Take the PHT (10aa 15aa 6aa; we have: 10 × 15 × 6 = 900 = 30�.

Four perfect squares associated with each primitive harmonic triple! Remarkable. But the claims are easier

to prove than one may expect. To do so we use the complementary factor algorithm obtained in Part–2 for

�inding such triples.

The algorithm recalled

For readers' convenience we derive the algorithm afresh. Suppose that aaaaaa aaaaaa aaaa are positive integers such

that aaaa aa aaaa and 1/aaaa aa 1/aaaa = 1/aaaa. By clearing fractions we get:

1

aaaa

aa

1

aaaa = 1

aaaa

aa ∴ aaaa aa aaaa

aaaaaaaa = 1

aaaa

aa ∴ aaaaaaaa aa aaaaaaaa = aaaaaaaaaa

hence aaaaaaaa aa aaaaaaaa aa aaaaaaaa = 0. Adding aaaa� to both sides we get:

aaaaaaaa aa aaaaaaaa aa aaaaaaaa aa aaaa� = aaaa�aa ∴ (aaaa aa aaaaaa(aaaa aa aaaaaa = aaaa�.

So aaaa aa aaaa and aaaa aa aaaa are a pair of divisors of aaaa� whose product is aaaa�. (Thus, they are a pair of ‘complementary

divisors’ of aaaa�.) So the algorithm to generate harmonic triples is:

1. Select a positive integer aaaa.

2. Write aaaa� as a product uuuu × uuuu of two positive integers, where uuuu aa uuuu.

3. Let aaaa = aaaa aa uuuu and aaaa a aaaa aa aaaa.

4. Then (aaaaaa aaaaaa aaaaaa = (aaaa aa uuuuaa aaaa aa uuuuaa aaaaaa is a harmonic triple in which aaaa aa aaaa.

In Part–2 of the article (July 2013 issue of At Right Angles) we remarked that to ensure that aaaaaa aaaaaa aaaa are

coprime (i.e., ensure that the triple is primitive), it is necessary as well as suf�icient that uuuu and uuuu be coprime.

Now we establish this claim.

Suppose that (aaaaaa aaaaaa aaaaaa = (aaaa aa uuuuaa aaaa aa uuuuaa aaaaaa is not primitive; then there exists a number kkkk kk 1 which divides

each number in the triple. Since kkkk divides both aaaa aa aaaa and aaaa, it divides uuuu. Since kkkk divides both aaaa aa aaaa and aaaa, it

divides uuuu. Therefore, kkkk divides both uuuu and uuuu. Hence uuuu and uuuu are not coprime. Taking the contrapositive of

this �inding, we deduce that if uuuu and uuuu are coprime, then the triple (aaaa aa aaaaaa aaaa aa aaaaaa aaaaaa is primitive.

Next we must show the converse: if uuuu and uuuu are not coprime then (aaaa aa aaaaaa aaaa aa aaaaaa aaaaaa is not primitive. Let pppp be

a prime number which divides uuuu as well as uuuu. Then pppp� is a divisor of aaaa�, since uuuuuuuu = aaaa�. Since pppp is prime, it

follows that pppp divides aaaa. Hence pppp divides each number in the triple (aaaa aa aaaaaa aaaa aa aaaaaa aaaaaa. Consequently the triple

is not primitive.

1

Page 3 of 5

22 At Right Angles | Vol. 3, No. 1, March 2014

Proof of Proposition 1

Let (aaaaaa aaaaaa aaaaaa be a PHT; then there exist coprime positive integers uuuu and vvvv such that uuuuvvvv uu aaaa�, aaaa uu aaaa aa uuuu,

aaaa aa aaaa aa aaaa. Since uuuu and vvvv are coprime and their product is a perfect square, each of them must be a perfect

square, say uuuu uu uuuu� and vvvv uu vvvv�; this yields aaaa aa aaaa uu uuuu� and aaaa aa aaaa aaaaaa�, showing directly that both aaaa aa aaaa and

aaaa aa aaaa are perfect squares, as claimed. Now consider aaaa aa aaaa. Since aaaa uu aaaa aa uuuu� and aaaa aa aaaa aa aaaa� and uuuuuuuu uu uuuu, we

have:

aaaa aa aaaa uu aaaa aa uuuu� aa aaaa aa

aaaa�

uuuu� uu uuuu� aa 2aaaa aa

aaaa�

uuuu�

uu uuuuu uu

aaaa

uuuu

uu (uuuu uuuuuuuu�.

So aaaa aa aaaa too is a perfect square.

Another pretty relation

We have shown that if (aaaaaa aaaaaa aaaaaa is a PHT, then

aaaa aa aaaa aa a√aaaa aa aaaa aa √aaaa aa aaaaa�

.

Example: Take the PHT (10aa 15aa 6aa; we have: 10 aa aaaaaaaa a√10 aa 6 aa √15 aa aa

.

Geometric interpretation of an algebraic relation

We now give a geometric interpretation to the following fact: If (aaaaaa aaaaaa aaaaaa is a harmonic triple then

(aaaa aa aaaaaa(aaaa aa aaaaaa uu aaaa�.

Given a △AAAAAAAAAAAA, let a rhombus AAAADDDDDDDDDDDD be inscribed in the triangle, with DDDD on AAAAAAAA, DDDD on AAAAAAAA, and DDDD on AAAAAAAA.

Figure 1 shows the completed picture. Let aaaaaa aaaaaa aaaa be the lengths indicated (aaaa uu uuuuuuuu, aaaa aaaaaaaaaa, aaaa aa side of the

rhombus). We had shown earlier that 1/aaaa aa 1/aaaa uu uuaaaa.

� �

� � � �

� � � �

� • �� � �

• �� � �

• ��, ��, ��, �� all have

length �

Figure 1: Rhombus inscribed in a triangle

Now note the similarity △AAAAAAAAAAAA BB △DDDDDDDDDDDD, which follows from the relations DDDDDDDD PP AAAAAAAA and DDDDDDDD DDDDDDDDDD. From

this we get the following relation:

aaaa aa aaaa

aaaa uu aaaa

aaaa aa aaaa .

By cross-multiplying we get the desired relation: (aaaa aa aaaaaa(aaaa aa aaaaaa uu aaaa�.

2