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20 At Right Angles | Vol. 3, No. 1, March 2014
Observe a relationship, then prove it – satisfying in itself.
Take this one step further and find the geometrical connect.
PHTs . . . Primitive and beautiful
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
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At Right Angles | Vol. 3, No. 1, March 2014 21
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:
aaaa = 1
aa ∴ aaaa aa aaaa
aaaaaaaa = 1
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.
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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
aaaa aa aaaa uu aaaa aa uuuu� aa aaaa aa
uuuu� uu uuuu� aa 2aaaa aa
uu uuuuu uu
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
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�.