Cassini's Identity/Proof 1

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Theorem

$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$


Proof

We see that:

$F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$

so the proposition holds for $n = 1$.

We also see that:

$F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$

so the proposition holds for $n = 2$.


Suppose the proposition is true for $n = k$, that is:

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

It remains to be shown that it follows from this that the proposition is true for $n = k + 1$, that is:

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


So:

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


By the Principle of Mathematical Induction, the proof is complete.

$\blacksquare$


Note that from the above we have that:

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

from which:

$F_{n + 1}^2 - F_n F_{n + 2} = \left({-1}\right)^n$

follows immediately.


Source of Name

This entry was named for Giovanni Domenico Cassini.


Sources

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