For an introduction to this series see this article. Essentially, I show examples of beautiful math that are accessible to people with limited math education.

**The Fibonacci numbers**

The Fibonacci sequence starts with two 1’s, then each subsequent number is the sum of the previous two. The sequence starts:

- 1, 1, 2, 3, 5, 8 ….

**The problem** I got this problem from James Tanton’s twitter feed (@JamesTanton). He tweets lots of interesting math problems. Here it is, as he tweeted it:

- Is the product of any k consecutive Fibonacci numbers always a multiple of the product of the first k Fibonacci numbers?

Before reading on, I urge you (very strongly!) to play around. Mathematics is all about play.

**My approach to a proof**

I have not come up with what I would consider a full, formal proof. But I think I am well on the way. It will be much easier to write about if you know about modular notations. Essentially, modular notation gives remainders from division. For instance 3 mod 2 is the remainder when 3 is divided by 2; that is, 1.

Sometimes when you see a problem, a path to a proof leaps out at you. But this didn’t happen in this case, at least not for me. So, I played. I tried multiplying the first few Fibonacci numbers, then seeing what I got. The first two are easy: 1*1 = 1, and every number is a product of 1. 1*1*2 = 2. Every even number is a product of 2 (by definition). Any three numbers in the Fibonacci sequence can be represented by a, b and a + b, Now, if either a or b is even, then the product of these three numbers must be even. But if both a and b are odd, then a + b must be even and, again, the product must be even. After that, it got a little trickier. However, I took the next step:

Let’s look at 4 terms: 1*1*2*3 = 6. Any four terms of the Fibonacci sequence are representable as a, b, a + b and a + 2b. Since any three terms are divisible by 2 (see above) all we need show is that the product of four terms is divisible by 3 (because if a number is divisible by 2 and by 3, it must be divisible by 6). In modular terms there are these possibilities:

- a mod 3 or b mod 3 = 0; that is, one of the two is divisible by 3. In that case the product of all 4 terms must be divisible by 3.
- a mod 3 = 1 and b mod 3 = 1. In this case, (a + 2b) is divisible by 3.
- a mod 3 = 2 and b mod 3 = 2. In this case (a + 2b) is divisible by 3.
- a mod 3 = 1 and b mod 3 = 2, or vice versa. In this case (a + b) is divisible by 3.

Next, I looked at 5 terms: 1*1*2*3*5 and the product of a, b, (a + b), (a + 2b) and (2a + 3b) and I noticed that the same thing happens, mod 5, as happened mod 3. And, although I haven’t proven it, it seems that the same thing happens with a sequence of any length.

A formalization of this proof might be by mathematical induction, but I haven’t worked it out completely.