120 degree and 60 degree triples and the divisors 3, 5, and 7.

This section in pdf form. triples357.pdf

It's well known that if $(a,b,c)$ is a a Pythagorean triple, that is, if $(a,b,c)$ is a solution in positive integers to the 90 degree triangle equation $a^2+b^2=c^2$ , then 3 and 4 each divides $a$ or $b$ , and 5 divides $a, b$ or $c$ where, of course, $(3,4,5)$ is the smallest such solution.

A 120 degree triple, $(a,b,c)$ , is a solution in positive integers to the 120 degree triangle equation

$$a^2+b^2-2ab\cos120 ^{\circ}=a^2+b^2+ab=c^2.$$

So, naturally, one wonders if a similar relationship exists between the positive integer solutions of 120 degree triangles and the smallest such solution, $(3,5,7)$ ? It does, almost. To find such a relationship it's necessary to include an $a+b$ term.

120 degree triples and the divisors 3, 5, and 7.

All primitive solutions to a 120 degree triple $(a,b,c)$ , are given by the parametric equations:

$$a=m^2-n^2,\quad b=2mn+n^2, \quad c=m^2+n^2+mn.$$ (37)

where $m$ and $n$ are relatively prime, positive integers, $m > n$ , and $3\nmid m-n$ . See (8) for a proof.

If $(a,b,c)$ and $(b,a,c)$ are considered the same solution, then the first 6 primitive solutions in order of smallest value for c are,

(1)

     $5^2+3^2+5\cdot 3=7^2$

(2)

     $8^2+7^2+8\cdot 7=13^2$

(3)

     $16^2+5^2+16\cdot 5=19^2$

(4)

     $24^2+11^2+24\cdot 11=31^2$

(5)

     $33^2+7^2+33\cdot 7=37^2$

(6)

     $35^2+13^2+35\cdot 13=43^2$

Notice that, in each case, 3 and 5, each, divides one of $a, b$ , or $a+b$ , and 7 divides one of $a, b$ , $a+b$ , or $c$ .

Claim 8   If $(a,b,c)$ is any 120 degree triple then 3 and 5 divides $ab(a+b)$ , and 7 divides $ab(a+b)c$ .

Proof. It's sufficient to show it's true for primitive triples. This claim can be proven directly by looking at residues modulo 3, 5, and 7; however it gets quite messy for the divisor 7. So, instead, I will use the parametric equations from (37) and the following result from Fermat's little theorem. That is, if $s$ and $t$ are integers, and $p$ is a prime, then

$$p$$   divides$$st\left(s^{p-1}-t^{p-1}\right).$$

From (37),

$$ab(a+b) =\left(m^2-n^2\right)\left(2mn+n^2\right)\left(m^2+2mn\right)$$

$$=m n\left(m^2 - n^2\right)\left(2\left(m^2 + n^2\right) - 5m n\right)$$

$$=2mn\left(m^4-n^4\right)-5\left(m^4n^2-m^2n^4\right) ,$$

and

$$ab(a+b)c =\left(m^2-n^2\right)\left(2mn+n^2\right)\left(m^2+2mn\right) \left(m^2+n^2+mn\right)$$

$$=2m n\left(m^6 - n^6\right) - 7\left(m^6 n^2 - m^5 n^3 + m^3 n^5 - m^2 n^6\right) .$$

Therefore, from Fermat's little theorem, 3 and 5 divide $ab(a+b)$ , and 7 divides $ab(a+b)c$ . And since 3, 5, and 7 are primes then 3 and 5 each divides one of $a, b$ , or $a+b$ , and 7 divides one of $a, b, a+b$ , or $c$ . $\qedsymbol$

60 degree triples and the divisors 3, 5, and 7.

A 60 degree triple, $(p,q,r)$ , is a solution in positive integers to the 60 degree triangle equation

$$p^2+q^2-2pq\cos60 ^{\circ}=p^2+q^2-pq=r^2.$$

Note that

$$a^2+b^2+ab=(a+b)^2+b^2-(a+b)b=a^2+(a+b)^2-a(a+b).$$

Hence, if $(a,b,c)$ is a 120 degree triple then $(a+b,b,c)$ and $(a,a+b,c)$ are 60 degree triples.

Here is a ``neat'' way to construct these three triangles.

On line $l$ layout line segments $AB$ and $BE$ having lengths $a$ and $b$ respectively, where $a$ and $b$ are the side lengths of a 120 degree triangle. On and below $AB$ construct equilateral triangle $ADB$ with sides of length $a$ . On and above $BE$ construct equilateral triangle $BEC$ with sides of length $b$ . Hence $\angle ABD$ and $\angle CBE$ are each 60 degrees. So point $B$ lies on line segment $DC$ and $\angle ABC$ is 120 degrees. Draw line segment $AC$ . Thus, the construction shows the 120 degree triangle $ABC$ and its two associated 60 degree triangles $AEC$ and $ADC$ .

Let $u^2+v^2-uv=w^2$ . If $u, v$ , and $w$ are positive integers, then $(u,v,w)$ is a 60 degree triple. If, additionally, $u, v$ , and $w$ are pairwise relatively prime, then $(u,v,w)$ is a primitive 60 degree triple. The first seven such triples in order of the smallest value for $w$ are,

(1)

     $1^2+1^2-1\cdot1= 1^2$

(2)

     $8^2+5^2-8\cdot5= 7^2$

(3)

     $8^2+3^2-8\cdot 3=7^2$

(4)

     $15^2+7^2-15\cdot 7=13^2$

(5)

     $15^2+8^2-15\cdot 8=13^2$

(6)

     $21^2+5^2-21\cdot 5=19^2$

(7)

     $21^2+16^2-21\cdot 16=19^2$

Notice that, in each case, 3 and 5, each, divides one of $u, v$ , or $u-v$ , and 7 divides one of $u, v$ , $u-v$ , or $w$ .

Claim 9   If $(u,v,w)$ is any 60 degree triple then 3 and 5 divides $uv(u-v)$ , and 7 divides $uv(u-v)w$ .

Proof. It's sufficient to show the claim is true for primitive triples. Clearly it's true for the triple $(1,1,1)$ . So let $u^2+v^2-uv=w^2$ where $(u,v,w)$ is a primitive triple, $uvw\neq 1$ . Without loss of generality, let $u$ be greater than $v$ , then

$$(u-v)^2+v^2+(u-v)v=u^2+v^2-uv=w^2.$$

Hence $(u-v,v,w)$ is a 120 degree triple. So, from claim (8),

$$3 \cdot 5 \mid (u-v)v\bigl((u-v)+v\bigr)=uv(u-v),$$   and$$\quad 7 \mid (u-v)v\bigl((u-v)+v\bigr)w=uv(u-v)w.$$

$\qedsymbol$

The drawing below shows two 60 degree triangles $AEC$ and $ADC$ along with their associated 120 degree triangle $ABC$ .

If the opposite side is to a power

Theorem 8   Let $a$ and $b$ be non-zero integers, and let $c, k,$ and $d$ be positive integers such that $a^2+b^2+ab=\left(c^k\right)^2$ where $d$ is a divisor of $k$ and $a, b,$ and $c$ are pairwise relatively prime.

1. If $p=6d-1$ is a prime then $p$ divides one of $a, b,$ and $a+b.$
2. If $q=6d+1$ is a prime then $q$ divides one of $a, b, a+b,$ and $c.$
3. If $d=3^j$ , $j$ a non-negative integer, then $3^{j+1}$ divides one of $a, b,$ and $a+b.$

To find relative prime non-zero integer solutions to $a^2+b^2+ab=\left(c^k\right)^2$ , set

$$a=\frac{\omega (m-n\bar{\omega})^{2k}-\bar{\omega}(m-n\omega)^{2k... ...\mbox{and}\quad b=\frac{ (m-n\bar{\omega})^{2k}-(m-n\omega)^{2k}}{\sqrt{3} i}$$

where

$$\omega=\frac{-1}{2}+\frac{1}{2} \sqrt{3} i,\quad \bar{\omega}=\frac{-1}{2}-\frac{1}{2} \sqrt{3} i,$$   and$$\quad i=\sqrt{-1} .$$

And where $m$ and $n$ are positive integers, $m > n$ , and $m-n$ is not a multiple of $3.$ Then $c=m^2+n^2+mn$ .

• If $a$ and $b$ have the same sign then $\left(\vert a\vert,\vert b\vert,c^k\right)$ is a 120 degree triple.
• If $a$ and $b$ have opposite sign then $\left(\vert a\vert,\vert b\vert,c^k\right)$ is a 60 degree triple.

Examples

Table 6: Divisors of 120 degree triples

Triangle	$k$	$d$	Divisors of $a, b,$ or $a+b$	Divisors of $a, b, a+b,$ or $c$
$a^2+b^2+ab=c^2$	1	1	$3, 5$	$3, 5, 7$
$a^2+b^2+ab=\left(c^2\right)^2$	2	1,2	$3, 5, 11$	$3, 5, 7, 11, 13$
$a^2+b^2+ab=\left(c^3\right)^2$	3	1,3	$5, 3^2, 17$	$5, 7, 3^2, 17, 19$
$a^2+b^2+ab=\left(c^4\right)^2$	4	1,2,4	$3, 5, 11, 23$	$3, 5, 7, 11, 13, 23$
$a^2+b^2+ab=\left(c^5\right)^2$	5	1,5	$3, 5, 29$	$3, 5, 7, 29, 31$
$a^2+b^2+ab=\left(c^6\right)^2$	6	1,2,3,6	$5, 3^2, 11, 17$	$5, 7, 3^2, 11, 13, 17, 19, 37$
$a^2+b^2+ab=\left(c^7\right)^2$	7	1,7	$3, 5, 41$	$3, 5, 7, 41, 43$
$a^2+b^2+ab=\left(c^8\right)^2$	8	1,2,4,8	$3, 5, 11, 23, 47$	$3, 5, 7, 11, 13, 23, 47$
$a^2+b^2+ab=\left(c^9\right)^2$	9	1,3,9	$5, 17, 3^3, 53$	$5, 7, 17, 19, 3^3, 53$
$a^2+b^2+ab=\left(c^{10}\right)^2$	10	1,2,5,10	$3, 5, 11, 29, 59$	$3, 5, 7, 11, 13, 29, 31,59, 61$

Note that if integers $a$ and $b$ have the same sign then $\vert a\vert+\vert b\vert$ and $a+b$ have the same divisors. And if $a$ and $b$ have opposite parity then $\vert a\vert+\vert b\vert$ and $a-b$ have the same divisors. So to change table (6) to a table of divisors of 60 degree triples all that's necessary is to change $a+b$ to $a-b.$

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