Theorem: Squares which are Sum of Two Cubes/Formulation 1

{{begin-eqn}}
{{eqn | l = A
      | r = 3 n^3 + 6 n^2 - n
}}
{{eqn | l = B
      | r = -3 n^3 + 6 n^2 + n
}}
{{eqn | l = C
      | r = 6 n^2 \paren {3 n^2 + 1}
}}
{{end-eqn}}
where $n \in \C$.

Proof:
Assume that:
:$A + B = 12 n^2$
Thus, factoring $A^3 + B^3$, we find that:
{{begin-eqn}}
{{eqn	| l = A^3 + B^3
	    | r = \paren {A + B} \paren {A^2 - A B + B^2}
	    | c = Sum of Two Cubes
}}
{{eqn	| ll = \leadsto
        | l = \frac {\paren {A^3 + B^3} } {\paren {A + B} }
        | r = \paren {A^2 - A B + B^2}
	    | c = 
}}
{{eqn	| ll = \leadsto
        | n = 1
        | l = \frac {C^2} {12 n^2}
        | r = \paren {A^2 - A B + B^2}
	    | c = $A^3 + B^3 = C^2$ and $A + B = 12 n^2$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn	| l = \paren {A - B}^2
	    | r = \paren {A^2 - 2 A B + B^2} 
	    | c = Square of Difference
}}
{{eqn	| r = \frac 4 3 \paren {A^2 - A B + B^2} - \frac 1 3 \paren {A^2 + 2 A B + B^2}
	    | c = 
}}
{{eqn	| r = \frac 4 3 \paren {\frac {C^2} {12 n^2} } - \frac 1 3 \paren {A + B }^2
	    | c = From $\paren 1 $ above, Square of Sum
}}
{{eqn	| r = \frac {C^2} {9 n^2} - \frac 1 3 \paren {12 n^2 }^2
	    | c = $A + B = 12 n^2$
}}
{{eqn	| ll = \leadsto
        | n = 2
        | l = \paren {A - B}^2
        | r = \frac {C^2} {9 n^2} - 48 n^4
	    | c = 
}}
{{end-eqn}}
Next, assume that:
:$C = n^2 \paren {\alpha n^2 + \beta}$
for some $\alpha, \beta \in \Z$
Furthermore, we would like to write:
:$\paren {A - B}^2 = \dfrac 1 9 n^2 \paren {\alpha n^2 - \beta}^2$
{{begin-eqn}}
{{eqn	| l = \frac 1 9 n^2 \paren {\alpha n^2 - \beta}^2
	    | r = \frac {C^2} {9 n^2} - 48 n^4
	    | c = from $\paren 2$ above
}}
{{eqn	| ll = \leadsto
        | l = \frac 1 9 n^2 \paren {\alpha^2 n^4 - 2 \alpha \beta n^2 + \beta^2}
        | r = \frac 1 {9 n^2} \paren {n^2 \paren {\alpha n^2 + \beta} }^2 - 48 n^4
	    | c = $C = n^2 \paren {\alpha n^2 + \beta}$
}}
{{eqn	| ll = \leadsto
        | l = \frac 1 9 \alpha^2 n^6 - \frac 2 9 \alpha \beta n^4 + \frac 1 9 \beta^2 n^2
        | r = \frac 1 {9 n^2} \paren {n^4 \paren {\alpha^2 n^4 + 2 \alpha \beta n^2 + \beta^2} } - 48 n^4
	    | c = 
}}
{{eqn	| ll = \leadsto
        | l = - \frac 2 9 \alpha \beta n^4
        | r = \frac 2 9 \alpha \beta n^4 - 48 n^4
	    | c = 
}}
{{eqn	| ll = \leadsto
        | l = 48
        | r = \frac 4 9 \alpha \beta
	    | c = 
}}
{{eqn	| ll = \leadsto
        | l = 108
        | r = \alpha \beta
	    | c = 
}}
{{end-eqn}}
Letting $\alpha = 18$, then $\beta = 6$ and $C = n^2 \paren {18 n^2 + 6} = 6 n^2 \paren {3 n^2 + 1}$
{{begin-eqn}}
{{eqn	| l = A + B
	    | r = 12 n^2
	    | c = 
}}
{{eqn	| l = A - B
        | r = \dfrac n 3 \paren {18 n^2 - 6}
	    | c = 
}}
{{eqn	| ll = \leadsto
        | l = 2 A
        | r = \dfrac n 3 \paren {18 n^2 - 6} + 12 n^2
	    | c = adding $2$ equations above
}}
{{eqn	| ll = \leadsto
        | l = A
        | r = 3 n^3 + 6 n^2 - n
	    | c = dividing by $2$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn	| l = A + B
	    | r = 12 n^2
	    | c = 
}}
{{eqn	| l = A - B
        | r = \dfrac n 3 \paren {18 n^2 - 6}
	    | c = 
}}
{{eqn	| ll = \leadsto
        | l = 2 B
        | r = - \dfrac n 3 \paren {18 n^2 - 6} + 12 n^2
	    | c = subtracting $2$ equations above
}}
{{eqn	| ll = \leadsto
        | l = B
        | r = -3 n^3 + 6 n^2 + n
	    | c = dividing by $2$
}}
{{end-eqn}}
{{qed}}