120 degree primitive triples,(a,b,c), where a - b=1.

$(a,b,c)$ is a primitive solution in positive integers to the 120 degree equation $a^2+b^2+ab=c^2$ if and only if there exists relatively prime, positive integers $m$ and $n$ , $m > n$ , $m\not\equiv n \pmod{3}$ such that

$$a=m^2-n^2,\quad b=2mn+n^2, \quad c=m^2+n^2+mn.$$ (38)

Finding solutions where a-b=1

(We'll need these first four recursive solutions to find the first matrix.)

Let $a-b=\left(m^2-n^2\right)-\left(2mn+n^2\right)=(m-n)^2-3n^2=1$ , a Pell equation. Hence, solutions are

$$m-n=\frac{(2+\sqrt{3})^s+(2-\sqrt{3})^s}{2}=f(s)$$   and$$\quad n=\frac{(2+\sqrt{3})^s-(2-\sqrt{3})^s}{2\sqrt{3}}=g(s).$$

Then $m=f(s)+g(s)$ and $n=g(s)$ . Thus

$$a= \bigl(f(s)+g(s)\bigr)^2-g(s)^2 ,$$

$$b= 2\bigl(f(s)+g(s)\bigr)g(s)+g(s)^2 ,$$

$$\quad c= \bigl(f(s)+g(s)\bigr)^2+g(s)^2+\bigl(f(s)+g(s)\bigr) g(s) .$$

Examples

Table 7: $a-b=1$ .

s	f(s)	g(s)	m	n	a	b	c
1	2	1	3	1	8	7	13
2	7	4	11	4	105	104	181
3	26	15	41	15	1456	1455	2521
4	97	56	153	56	20273	20272	35113

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