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**Archie Meade** Here is another way;

**P(k)**

$\displaystyle F_{k+2}\;F_k-\left(F_{k+1}\right)^2=(-1)^k$

**P(k+1)**

$\displaystyle F_{k+3}\;F_{k+1}-\left(F_{k+2}\right)^2=(-1)^{k+1}=(-1)(-1)^k$

**Proof**

Try to show that P(k+1) will be true if P(k) is true,

hence write P(k+1) in terms of P(k).

$\displaystyle F_{k+3}=F_{k+1}+F_{k+2}$

$\displaystyle F_{k+2}=F_{k+1}+F_k}$

therefore, rewriting P(k+1)

$\displaystyle \left(F_{k+1}+F_{k+2}\right)F_{k+1}-\left(F_{k+1}+F_k\right)\left(F_{k+1}+F_k\right)=-(-1)^k$ ?

$\displaystyle \left(F_{k+1}\right)^2+F_{k+2}\;F_{k+1}-\left[\left(F_{k+1}\right)^2+2F_k\;F_{k+1}+\left(F_k\rig ht)^2\right]=-(-1)^k$ ?

$\displaystyle F_{k+2}\;F_{k+1}-2F_k\;F_{k+1}-\left(F_k\right)^2=-(-1)^k$ ?

$\displaystyle \left(F_k\right)^2+2F_k\;F_{k+1}-F_{k+2}\;F_{k+1}=(-1)^k$ ?

If we now write $\displaystyle F_{k+2}$ in terms of $\displaystyle F_{k+1},$ we will obtain the $\displaystyle \left(F_{k+1}\right)^2$ term in P(k)...

$\displaystyle F_{k+1}+F_k=F_{k+2}$

$\displaystyle \Rightarrow\ \left(F_k\right)^2+2F_k\;F_{k+1}-\left(F_k+F_{k+1}\right)F_{k+1}=\left(F_k\right)^2 +2F_k\;F_{k+1}-F_k\;F_{k+1}-\left(F_{k+1}\right)^2$

$\displaystyle =\left(F_k\right)^2+F_k\;F_{k+1}-\left(F_{k+1}\right)^2$

Factoring out $\displaystyle F_k$

$\displaystyle F_k\left(F_k+F_{k+1}\right)-\left(F_{k+1}\right)^2=F_k\;F_{k+2}-\left(F_{k+1}\right)^2$

Hence if P(k) is true, P(k+1) will certainly be true.

Test the base case.