Proof. According to Defintion 1 and Theorem 1, a non-equilateral primitive Eisensteinian triangle is an integer triangle with side lengths (a,b,c) such that the angle opposite the side of length a is of \(60^{\circ}\) and gcd(a,b,c)=1 and b\({>}\)c.

Now we use a 3\(\times\)​3 matrix \[\rm S = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right)\] and present above-mentioned (a,b,c) as a column, then the product of the two is: \[\rm S \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right) = \left( \begin{array}{c} \rm a \\ \rm c \\ \rm b-c \end{array} \right)\]

Now we assume that (a,b,c) is an element the set of non-equilateral primitive Eisensteinian triangles.

By Definition 1, the Law of Cosines says that:
(2.1) \({\rm a^2=b^2+c^2-2bc\;cos60^\circ=b^2+c^2-bc}\)
This is followed by:
(2.2) \({\rm a^2=c^2+(b-c)^2+c(b-c)=c^2+(b-c)^2-2c(b-c)\;cos120^\circ}\)
According to Definition 2, (a,c,b\({-}\)c) is an element of the set of primitive sub-Eisensteinian triangles, because b\({-}\)c is a positive integer and gcd(a,c,b\({-}\)c)=1, owing to gcd(a,b,c)=1.
Inversely we assume that (a,b,c) is an element the set of primitive sub-Eisensteinian triangles. Now we compute the inverse of S: \[\rm S^{-1} = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right)\]
and present above-mentioned (a,b,c) as a column, then the product of the two is: \[\rm S^{-1} \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right) \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right) = \left( \begin{array}{c} \rm a \\ \rm b+c \\ \rm b \end{array} \right)\]

Now we assume that (a,b,c) is an element the set of primitive sub-Eisensteinian triangles.

By Definition 2, the Law of Cosines says that
(2.3) \({\rm a^2=b^2+c^2-2bc\;cos120^\circ=b^2+c^2+bc}\)
This is followed by
(2.4) \({\rm a^2=(b+c)^2+b^2-(b+c)b=(b+c)^2+b^2-2(b+c)b\;cos60^\circ}\)
According to Definition 1, (a,b\({+}\)c,b) is an element of the set of non-equilateral primitive Eisensteinian triangles , because b\({+}\)c\({>}\)b and gcd(a,b\({+}\)c,b)=1, owing to gcd(a,b,c)=1.
These complete the proof. ◻

Proof. By using S, we see \[\rm S \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm m^2{+}2mn \\ \rm n^2{+}2mn \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm m^2{+}2mn \\ \rm n^2{+}2mn \end{array} \right) = \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm n^2{+}2mn \\ \rm m^2{-}n^2 \end{array} \right) \;and\] \[\rm S \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm m^2{+}2mn \\ \rm m^2{-}n^2 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm m^2{+}2mn \\ \rm m^2{-}n^2 \end{array} \right) = \left( \begin{array}{c} \rm m^2{+}mn{+}n^2 \\ \rm m^2{-}n^2 \\ \rm n^2{+}2mn \end{array} \right) .\]

According to what Theorem 3 says, this completes Corollary 1. ◻

Proof. According to Theorem 1, a non-equilateral integer sided triangle is primitive Eisensteinian if, and only if it has side lengths (a,b,c) or (a,b,b\({-}\)c) such that: \[\rm \left(\rm \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right) = M\cdot v,\;where\;v{\in} \left\{ \left( \begin{array}{c} 7 \\ 8 \\ 5 \end{array} \right){,} \left( \begin{array}{c} 13 \\ 15 \\ 17 \end{array} \right) \right\}\]
And M is a finite (or empty) product of the matrices: \[\rm M_1=\left( \begin{array}{ccc} 7 & -6 & 6 \\ 8 & -7 & 7 \\ 4 & -4 & 3 \end{array} \right){,} M_2=\left( \begin{array}{ccc} 7 & 6 & -6 \\ 8 & 7 & -7 \\ 4 & 3 & -4 \end{array} \right){,} M_3=\left( \begin{array}{ccc} 7 & 6 & 0 \\ 8 & 7 & 0 \\ 4 & 3 & 1 \end{array} \right){,}\] \[\rm M_4=\left( \begin{array}{ccc} 7 & -6 & 6 \\ 8 & -7 & 7 \\ 4 & -4 & 3 \end{array} \right){,} M_5=\left( \begin{array}{ccc} 7 & 6 & -6 \\ 8 & 7 & -7 \\ 4 & 3 & -4 \end{array} \right)\]
and any such triangle is obtained exactly once.

We now temporally consider the case of side lengths (a,b,c).

Then we set
(2.5) M=\({\rm M_{p1}}\) \({\rm M_{p2}}\) \(\cdots\) \({\rm M_{pn}}\)

where for any integer k(1\(\leq\)k\(\leq\)n) and

\({\rm M_{pk}}\) = \({\rm M_1}\) or \({\rm M_2}\) or \({\rm M_3}\) or \({\rm M_4}\) or \({\rm M_5}\).
Moreover we obtain that:
(2.6) S\(\cdot\)M\(\cdot\)v=S(\({\rm M_{p1}}\) \({\rm M_{p2}}\) \(\cdots\) \({\rm M_{pn}}\))v

=(\({\rm SM_{p1}S^{-1}}\))(\({\rm SM_{p2}S^{-1}}\))\(\cdots\)(\({\rm SM_{pn}S^{-1}}\))(S\(\cdot\)v)
According to the proof of Theorem 2, S\(\cdot\)v is an element of the set of primitive sub-Eisensteinian triangles. And we easily see that two different elements of the set of primitive Eisensteinian triangles is respectively mapped to different elements of the set of primitive Eisensteinian triangles by S.

And owing to Theorem 1, if any two certificates for M that have a different expression from each other, then their values are also different from each other.

We arrange (2.6) as follows:
(2.7) S\(\cdot\)M\(\cdot\)v=S(\({\rm M_{p1}}\) \({\rm M_{p2}}\) \(\cdots\) \({\rm M_{pn}}\))v

=(\({\rm SM_{p1}S^{-1}}\))(\({\rm SM_{p2}S^{-1}}\))\(\cdots\)(\({\rm SM_{pn}S^{-1}}\))(S\(\cdot\)v)

=(\({\rm N_{p1}}\))(\({\rm N_{p2}}\))\(\cdots\)(\({\rm N_{pn}}\))(S\(\cdot\)v)

=\({\rm N}\)(S\(\cdot\)v).
We let:
(2.8) N=\({\rm N_{p1}}\) \({\rm N_{p2}}\) \(\cdots\) \({\rm N_{pn}}\)

where for any integer k(1\(\leq\)k\(\leq\)n) and

\({\rm N_{pk}}\) = \({\rm SM_{1}S^{-1}}\) or \({\rm SM_{2}S^{-1}}\) or \({\rm SM_{3}S^{-1}}\) or \({\rm SM_{4}S^{-1}}\) or \({\rm SM_{5}S^{-1}}\).
In the above-mentioned equations, any two certificates for N that have a different expression from each other, then their values are also different from each other.

Then by computing we see:

\[\rm N_1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 7 & -6 & 6 \\ 8 & -7 & 7 \\ 4 & -4 & 3 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right){=} \left( \begin{array}{ccc} 7 & 0 & -6 \\ 4 & -1 & -4 \\ 4 & 1 & -3 \end{array} \right){,}\] \[\rm N_2=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 7 & 6 & -6 \\ 8 & 7 & -7 \\ 4 & 3 & -4 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right){=} \left( \begin{array}{ccc} 7 & 0 & 6 \\ 4 & -1 & 3 \\ 4 & 1 & 4 \end{array} \right){,}\] \[\rm N_3=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 7 & 6 & 0 \\ 8 & 7 & 0 \\ 4 & 3 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right){=} \left( \begin{array}{ccc} 7 & 6 & 6 \\ 4 & 4 & 3 \\ 4 & 3 & 4 \end{array} \right){,}\] \[\rm N_4=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 7 & 0 & -6 \\ 8 & 0 & -7 \\ 4 & 1 & 3 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right){=} \left( \begin{array}{ccc} 7 & 6 & 0 \\ 4 & 4 & 1 \\ 4 & 3 & -1 \end{array} \right){,}\] \[\rm N_5=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{ccc} 7 & 0 & -6 \\ 8 & 0 & -7 \\ 4 & 1 & -4 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array} \right){=} \left( \begin{array}{ccc} 7 & -6 & 0 \\ 4 & -3 & 1 \\ 4 & -4 & -1 \end{array} \right),\] \[\rm Sv= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} 7 \\ 8 \\ 5 \end{array} \right) {=} \left( \begin{array}{c} 7 \\ 5 \\ 3 \end{array} \right) \;\;or\] \[\rm \;\;\;= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} 13 \\ 15 \\ 7 \end{array} \right) {=} \left( \begin{array}{c} 13 \\ 7 \\ 8 \end{array} \right).\]

The general expression of the elements of the set of primitive sub-Eisensteinian triangles, mapped by S from the elements of the set of non-equilateral primitive Eisensteinian triangles, in the case of side lengths (a,b,c), is (2.7).

Then next we consider the case of side lengths (a,b,b\({-}\)c). In this connection, the pair (a,b,c) and (a,b,b\({-}\)c) is called the the twin Eisensteinian triplet [1].

With the aid of Theorem 1, we obtain: \[\rm (2.9)\;\;S\left( \begin{array}{c} \rm a \\ \rm b \\ \rm b-c \end{array} \right)= S\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right)= S\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right)(M\cdot v)\] \[\rm =S\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{array} \right)S^{-1}(SMv)= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)(SMv)\]

It is evident that: \[\rm (2.10) \;\;\;\;\;\;\; \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right) \left( \begin{array}{c} \rm a \\ \rm b \\ \rm c \end{array} \right)= \left( \begin{array}{c} \rm a \\ \rm c \\ \rm b \end{array} \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\] therefore we see as follows:

If the case of side lengths (a,b,c) for primitive Eisensteinian corresponds to the case of side lengths (a,b,c) for primitive sub-Eisensteinian, then the case of side lengths (a,b,b\({-}\)c) for primitive Eisensteinian corresponds to the case of side lengths (a,c,b) for primitive sub-Eisensteinian.

Thus any primitive sub-Eisensteinian triangle is also obtained by pre-multiplication (7,5,3) or (13,7,8) by a product of five matrices.

These complete the proof. ◻