proof of Catalan’s Identity

proof of Catalan’s Identity

For all

positive integers

i

, let

F

i

denote the

i

t

⁢

h

Fibonacci number

, with

F

1

=

F

2

= 1. We will show that for all positive integers

n

and

r

such that

n

>

r

the following holds:

F

n

2

-

F

n

+

r

⁢

F

n

-

r

=

(

-

1

)

n

-

r

⁢

F

r

2

.

But in order to prove this we first need two lemmas:

Lemma 0.1

.

For all positive integers

a

and

b

such that

a

>

1

, the

identity

F

a

+

b

=

F

a

⁢

F

b

+

1

+

F

a

-

1

⁢

F

b

is true.

Proof.

We will prove this by

induction

on

b

. When

b

=

1

, the identity states that

F

a

+

1

=

F

a

⁢

F

2

+

F

a

-

1

⁢

F

1

⟺

F

a

+

1

=

F

a

⋅

1

+

F

a

-

1

⋅

1

which is true by the

definition

of the Fibonacci numbers. Now, assume for possible values of

b

less than some positive integer

b

0

such that

b

0

>

1

, the

proposition

is true. Then

F

a

⁢

F

b

0

+

1

+

F

a

-

1

⁢

F

b

0

=

F

a

⁢

(

F

b

0

+

F

b

0

-

1

)

+

F

a

-

1

⁢

(

F

b

0

-

1

+

F

b

0

-

2

)

=

(

F

a

⁢

F

b

0

+

F

a

-

1

⁢

F

b

0

-

1

)

+

(

F

a

⁢

F

b

0

-

1

+

F

a

-

1

⁢

F

b

0

-

2

)

=

F

a

+

(

b

0

-

1

)

+

F

(

a

-

1

)

+

(

b

0

-

1

)

(

induction hypothesis

)

=

F

a

+

b

0

(definition of the Fibonacci numbers)

This concludes the proof. ∎

Lemma 0.2

.

For all positive integers

t

such that

t

>

1

, the following holds:

F

t

-

1

2

+

F

t

⁢

F

t

-

1

-

F

t

2

=

(

-

1

)

t

.

Proof.

We will (again) proceed by induction. First, when

t

=

2

, we have

F

1

2

+

F

2

⁢

F

1

-

F

2

2

=

1

⟺

1

+

1

⋅

1

-

1

2

=

1

which is true. Now let us assume that the proposition is true for
all positive integers which are greater than 1 and less than some
positive integer

t

0

(

t

0

>

2

)

. Then

F

t

0

-

1

2

+

F

t

0

⁢

F

t

0

-

1

-

F

t

0

2

=

F

t

0

-

1

2

+

(

F

t

0

-

1

+

F

t

0

-

2

)

⁢

F

t

0

-

1

-

(

F

t

0

-

1

+

F

t

0

-

2

)

2

=

F

t

0

-

1

2

+

F

t

0

-

1

2

+

F

t

0

-

2

⁢

F

t

0

-

1

-

F

t

0

-

1

2

-

F

t

0

-

2

2

-

2

⁢

F

t

0

-

1

⁢

F

t

0

-

2

=

F

t

0

-

1

2

-

2

⁢

F

t

0

-

1

⁢

F

t

0

-

2

-

F

t

0

-

2

2

=

-

(

F

t

0

-

2

2

-

2

⁢

F

t

0

-

2

⁢

F

t

0

-

1

-

F

t

0

-

1

2

)

=

(

-

1

)

⁢

(

-

1

)

t

0

-

1

(by induction hypothesis)

=

(

-

1

)

t

0

and the proof is

complete

.
∎

Now to the main proposition. Let us make the substitutions

x

=

n

-

r

and

a

=

r

so that the

theorem

now states:

Theorem 0.3

.

For all positive integers

x

and

a

, the following identity holds:

F

x

+

a

2

-

F

x

+

2

⁢

a

⁢

F

x

=

(

-

1

)

x

⁢

F

a

2

.

Proof.

We follow a series of calculations

F

x

+

a

2

-

F

x

+

2

⁢

a

⁢

F

x

=

(

F

x

⁢

F

a

+

1

+

F

x

-

1

⁢

F

a

)

2

-

(

F

x

⁢

F

2

⁢

a

+

1

+

F

x

-

1

⁢

F

2

⁢

a

)

⁢

F

x

(by lemma 1)

=

F

x

2

⁢

F

a

+

1

2

+

2

⁢

F

x

⁢

F

a

+

1

⁢

F

x

-

1

⁢

F

a

+

F

x

-

1

2

⁢

F

a

2

-

⁢

F

x

⁢

(

F

x

⁢

(

F

a

+

1

2

+

F

a

2

)

+

F

x

-

1

⁢

(

F

a

⁢

F

a

+

1

+

F

a

-

1

⁢

F

a

)

)

(by lemma 1 again)

=

F

x

⁢

F

x

-

1

⁢

F

a

⁢

(

F

a

+

1

-

F

a

-

1

)

+

F

a

2

⁢

(

F

x

-

1

2

-

F

x

2

)

=

F

x

⁢

F

x

-

1

⁢

F

a

⁢

(

F

a

)

+

F

a

2

⁢

(

F

x

-

1

2

-

F

x

2

)

(by the definition of Fibonacci numbers)

=

F

a

2

⁢

(

F

x

-

1

2

+

F

x

⁢

F

x

-

1

-

F

x

2

)

=

F

a

2

⁢

(

-

1

)

x

(by lemma 2)

completing our proof of the theorem.
∎

Title

proof of Catalan’s Identity

Canonical

name

ProofOfCatalansIdentity

Date of creation

2013-03-22 14:45:01

Last modified on

2013-03-22 14:45:01

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

20

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11B39

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