# Fibs

## Problem

The Fibonacci sequence is

$1, 1, 2, 3, 5, 8, 13, 21 \ldots $

where each term is the sum of the two terms that go before it (i.e $1+1=2$, $1+2=3$, $2+3=5$ and so on.)

What is the sixth term of the Fibonacci type sequence that starts with $2$ and $38$ as the first two terms?

How many Fibonacci type sequences can you find containing the number $196$ as one of the terms where the sequence starts with two whole numbers $a$ and $b$ with $a< b$?

NOTES AND BACKGROUND

Fibonacci sequences are named after a merchant, one Leonardo of Pisa who had the nickname Fibonacci. On his travels, around 1200 AD, he learnt a lot of mathematics (particularly algebra) from the Arabs.

The Arabs had developed the study of mathematics for about 800 years after the fall of the Greek and Roman civilisations. The story behind the methods in this problem spans this whole period.

Equations in which one seeks whole number solutions, are called Diophantine equations after Diophantus (250 A.D) who developed the method and a special notation for recording it.

For more about the stories of Diophantus and Fibonacci see the History of Maths website .

## Student Solutions

Lots of people sent in the solution that
the sixth term of the Fibonacci sequence starting with $2$ and $38$
is $196$ and you found other sequences with $196$ as one of the
terms. Exactly how many other Fibonacci sequences contain the term
$196$? *A
lot of solutions as Jimmy rightly pointed out!*

The simplest Fibonacci sequence is:

$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 253, \ldots$

and we denote the $n$th term of this sequence by $F(n)$.

Starting with the terms $a, b$ (for $a$ and $b$ positive whole
numbers and $a < b$) we get the general Fib sequence:

$a$, $b$, $a+b$, $a+2b$, $2a+3b$, $3a+5b$, $5a+8b$, $8a+13b$,
$13a+21b$, $21a+34b$, $34a+55b$, $55a+89b$, $89a+144b$,
$\ldots$

The $n$th term of the general Fib sequence $f(n) = aF(n-2) + bF(n-1)$ and note that, if the term $196$ occurs in the sequence, it can't be beyond the twelfth term as after that the terms are too large.

Here are some sequences containing $196$.

**Sequences with $196$ as the first term**

$196, b, 196+b, 196+2b, \ldots$ | for $b > 196$ |

**Sequences with $196$ as the second term**

$a, 196, \ldots$ | for $1 < a < 195$ |

**Sequences with 196 as the third term**

$1, 195, 196, \ldots$

$2, 194, 196, \ldots$

$...$

$97, 99, 196, \ldots$

$98, 98, 196, \ldots$ etc

So far we see that there are infinitely many sequences with $196$
as the first term; exactly $195$ with $196$ as the second term;
exactly $98$ with $196$ as the third term.

To find all the remaining sequences containing $196$ we have to
find whole numbers $a$ and $b$ where:

$a + 2b = 196$

or

$2a + 3b = 196$

or

$3a + 5b = 196$

etc.

In general we have to find whole number values of $a$ and $b$
satisfying

$aF(n-2) + bF(n-1) = 196$.

and so we need to find whole number solutions to these equations
for $n = 4, 5, 6, \ldots12$.

We shall consider one remaining case and leave the rest to the
reader.

For $n = 6$ we seek values of $a$ and $b$ such that $3a + 5b = 196$

There are no solutions for $a = 1$ because then b would not be a
whole number. We have already seen that $a = 2$ and $b = 38$ gives
$196$ as the sixth term. For larger values of $a$ we have to take
smaller values of $b$. For $a = 3$ or $4$ or $5$ or $6$ there are
again no solutions because b has to be a whole number.

For $a = 7$ we have:

$21 + 5b$ | $=$ | $196$ |

$5b$ | $=$ | $175$ |

$b$ | $=$ | $35$ |

giving the sequence $7, 35, 42, 77, 119, 196, \ldots$

To find the remaining solutions for $n = 6$ we increase $a$ by
steps of $5$ and decrease $b$ by steps of $3$.

There are five solutions for $n = 6$, which are:

- $2, 38, \ldots$
- $7, 35, \ldots$
- $12, 32, \ldots$
- $17, 29,\ldots$
- $22, 26,\ldots$

You can use the same method to find the solutions for $n = 4, 5, 7, \ldots 12$

## Teachers' Resources

Fibonacci sequences are named after a merchant, one Leonardo of Pisa who had the nickname Fibonacci. On his travels, around 1200 AD, he learnt a lot of mathematics (particularly algebra) from the Arabs.

The Arabs had developed the study of mathematics for about 800 years after the fall of the Greek and Roman civilisations. The story behind the methods in this problem spans this whole period.

Equations in which one seeks whole number solutions, are called Diophantine equations after Diophantus (250 A.D) who developed the method and a special notation for recording it.

For more about the stories of Diophantus and Fibonacci see the History of Maths website .