Because determinants play nice with multiplication, we can take the determinant of both sides in our equation Mn=(F(n+1)F(n)F(n)F(n−1))M^n = \begin{pmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{pmatrix} and simplify, as I've done opposite. This gives us the result known as Cassini's identity.

We had M2n=(a+bbba)(a+bbba)=((a+b)2+b22ab+b22ab+b2a2+b2)=((a+b)2+b2(a+b)2−a2(a+b)2−a2a2+b2) \begin{aligned} M^{2n} &= \begin{pmatrix} a + b & b \\ b & a \end{pmatrix} \begin{pmatrix} a + b & b \\ b & a \end{pmatrix} \\ &= \begin{pmatrix} (a + b)^2 + b^2 & 2ab + b^2 \\ 2ab + b^2 & a^2 + b^2 \end{pmatrix} \\ &= \begin{pmatrix} (a + b)^2 + b^2 & (a + b)^2 - a^2 \\ (a + b)^2 - a^2 & a^2 + b^2 \end{pmatrix} \end{aligned}

That the bottom right corner of this is a2+b2a^2 + b^2 can be written as such: F(2n−1)=F(n)2+F(n−1)2F(2n - 1) = F(n)^2 + F(n - 1)^2. Can we use Cassini's identity to rewrite one of the other cells (F(2n)F(2n) or F(2n+1)F(2n + 1)) in a form that only requires computing F(n)2F(n)^2 and F(n−1)2F(n - 1)^2, not F(n+1)2F(n + 1)^2?

Because determinants play nice with multiplication, we can take the determinant of both sides in our equation Mn=(F(n+1)F(n)F(n)F(n−1))M^n = \begin{pmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{pmatrix} and simplify, as I've done opposite. This gives us the result known as Cassini's identity.

We had M2n=(a+bbba)(a+bbba)=((a+b)2+b22ab+b22ab+b2a2+b2)=((a+b)2+b2(a+b)2−a2(a+b)2−a2a2+b2) \begin{aligned} M^{2n} &= \begin{pmatrix} a + b & b \\ b & a \end{pmatrix} \begin{pmatrix} a + b & b \\ b & a \end{pmatrix} \\ &= \begin{pmatrix} (a + b)^2 + b^2 & 2ab + b^2 \\ 2ab + b^2 & a^2 + b^2 \end{pmatrix} \\ &= \begin{pmatrix} (a + b)^2 + b^2 & (a + b)^2 - a^2 \\ (a + b)^2 - a^2 & a^2 + b^2 \end{pmatrix} \end{aligned}

That the bottom right corner of this is a2+b2a^2 + b^2 can be written as such: F(2n−1)=F(n)2+F(n−1)2F(2n - 1) = F(n)^2 + F(n - 1)^2. Can we use Cassini's identity to rewrite one of the other cells (F(2n)F(2n) or F(2n+1)F(2n + 1)) in a form that only requires computing F(n)2F(n)^2 and F(n−1)2F(n - 1)^2, not F(n+1)2F(n + 1)^2?