#186: Math on Monday (7)

Chow Kim Wan

8 July 2015

I have decided to stick with this font for the Math on Monday posts, until furthernotice. Computer Modern is kind of too mainstream. Please tell me if this fontirritates you.

At first, I wanted to talk about complex numbers today, but a recent post by abyss-brain from the blog Mathemagical Site (link provided in my main blog post for this)urged me to share about this instead. Essentially, he shared about a pattern inthe Fibonacci series, which has been featured in this series several times already.Observe the following pattern:

3 consecutive terms Square of 2nd term Product of 1st and 3rd terms1, 1, 2 1 21, 2, 3 4 32, 3, 5 9 103, 5, 8 25 245, 8, 13 64 658, 13, 21 169 16813, 21, 34 441 442

The square of the second of any three consecutive terms in the Fibonacci sequenceis always either 1 greater or 1 less than the product of the first and third terms.To put it in another way, the square of any term in the Fibonacci sequence is onemore or one less than the product of the term before it and the one after it. This iswell-illustrated in the above table.

What we seek is a proof or justification of this assertion. In math, we cannot simplylist the first few iterations, and assume it will hold true for the general case. All wecan obtain from this is a conjecture. A conjecture is basically a fancy name for astatement which seems to be true but has not yet been proven. Famous conjectures inmathematics include Goldbach’s conjecture. For many years, Fermat’s Last Theoremstayed a conjecture, until Andrew Wiles proved it in 1995.

So, we shall conjecture right now that the square of any term in the Fibonaccisequence is one more or one less than the product of the term before it and theone after it. Our aim is to prove this conjecture, and once we manage to do so, itbecomes a theorem (albeit a quite useless one).

This is going to be straightforward once we have a closed form expression for the

1

nth term of the sequence (which shall be denoted as Fn). This has, in fact, been thetopic of discussion for many of our Math on Monday posts, and the formula andits various justifications can be found there. I shall simply quote the formula herewithout proof:

Fn =1√5

[(1+

√5

2

)n

−

(1−

√5

2

)n](1)

With this, we shall just meddle with the algebra, and evaluate F2n and Fn−1Fn+1. Toavoid having to write the surds too many times, we shall let t1 =

12

(1+

√5)

and

t2 = 12

(1−

√5). We should note at this point that t1 and t2 are the roots of the

equation t2 − t − 1 = 0, so that t1 + t2 = 1 and t1t2 = −1 by Vieta’s formulas. Fromthis,

t21 + t22 = (t1 + t2)2 − 2t1t2 = 3

This will come in handy later.

Whacking the expression for F2n,

F2n =15[tn1 − tn2 ]

2

=15[t2n1 + t2n2 − 2tn1 tn2

]=

15[t2n1 + t2n2 − 2 (−1)n

]Similarly, we have:

Fn−1Fn+1 =15[tn−11 − tn−1

2] [tn+11 − tn+1

2]

=15[t2n1 + t2n2 − tn−1

1 tn+12 − tn+1

1 tn−12]

=15[t2n1 + t2n2 − tn−1

1 tn−12(t21 + t22

)]=

15

[t2n1 + t2n2 − 3 (−1)n−1

]The difference can then be easily found, and our result immediately follows:∣∣Fn−1Fn+1 − F2n

∣∣ = 15

∣∣∣−3 (−1)n−1 + 2 (−1)n∣∣∣

=

∣∣∣∣(−1)n

5(3+ 2)

∣∣∣∣= 1

This is a simple verification of the observation using the closed form expressionof the Fibonacci sequence. Of course, one can dabble with the recurrence relationwithout invoking the closed form expression to obtain a similar result, and that canbe left as an exercise to the reader. Shall see everyone with complex numbers twoMondays later!

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