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Note: (written November 6 2016) I'm glad you found this essay I wrote long ago. It's archived here on my website, together with my other essays from a decade ago. If you want to see my more recent content, my blog is the place to find it.

The identity 3*5 = 15 in different contexts

by Walter Vannini

Introduction
Base 2 and an algebraic identity
Base 4 and and difference of squares
Base 2 and another algebraic identity
Triangular Numbers
Fibonacci numbers
Sums of Fibonacci squares
Three Times Generalized Fibonacci
The Golden Ratio
Mersenne primes
Product of Sum of Squares
Complex Numbers
Quaternions
An infinite product
Concluding words
Feedback

Introduction

It's often been said that all of mathematics is connected. This essay is about some of the mathematics that is very closely connected to the result "3*5=15". Amongst other things, that includes various algebraic identities, Fibonacci numbers, the Golden Ratio, Mersenne primes, sums of squares, complex numbers, and quaternions.

Base 2 and an algebraic identity

By viewing 3*5=15 in base 2, it's 11*101=1111. This tells us that the identity can be viewed as (2+1)(4+1)=8+4+2+1, which is the identity resulting from substituting x=2 in the algebraic identity

(x+1)(x2+1)=x3+x2+x+1.

Base 4 and difference of squares

Writing the identity in base 4, it's 3*11 = 33. In base 4, 3 is not just 1+1+1, but it is also the largest one digit number. It corresponds to 9 in base 10, so that the identity is really analogous to 9*11= 99.

The general equation is

(x-1)(x+1) = x2-1.

Base 2 and another algebraic identity

Writing 3*5=15 in base 2 as 11*101=1111, the fact that 1 is the largest digit in base 2 means that there are analogues in base 10 where 1 is replaced by 9. They are:

99*101 = 9999 and 11*909=9999 (I'm going to ignore 11*901=9911, 19*101=1199 etc)

both of which are really

9 * 11 * 101 = 9999

The underlying algebraic identity is

(x-1) (x+1) (x2+1) = x4 - 1

Triangular Numbers

15 is the triangular number (1+2+3+4+5).

3*5 = 15 can be viewed as the result of substituting n=5 in the the formula

(n+1)/2 * n = (1+2+3+…+n)

Fibonacci numbers

Recall that the Fibonacci sequence has terms

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

and can be defined recursively via:

Fn = Fn-1 + Fn-2
F0 = 0
F1 = 1

The fact that 3 and 5 are in the sequence suggests that 3*5 = 15 is a special case of an identity involving Fibonacci numbers.

Sums of Fibonacci squares

It turns out that the product of the nth Fibonacci number with the following Fibonacci number is the sum of the squares of the first n Fibonacci numbers. ie.

Fn * Fn+1 = F12 + F22 + … + Fn2.

Substituting the value n=4 in the above identity, we get

F4 * F5 = F12 + F22 + F32 + F42.

3 * 5 = 12 + 12 + 22 + 32.

Three Times Generalized Fibonacci

For a Generalized Fibonacci sequence (ie a sequence satisfying the Fibonacci recurrence) we have

3 * Gn = Gn+2+Gn-2

The proof is straightforward:

3Gn = Gn + Gn + Gn
= Gn + Gn + (Gn-1 + Gn-2)
= Gn + (Gn + Gn-1) + Gn-2
= Gn + Gn+1 + Gn-2
= (Gn + Gn+1) + Gn-2
= Gn+2 + Gn-2

Letting Gn be the Fibonacci sequence, and substituting n=5, we have

3 * F5 = F7 + F3

3 * 5 = 13 + 2.

The Golden Ratio

Recall that the golden ratio φ is (1+√5) ⁄ 2.

An exact formula for the Fibonacci numbers is given by

Fn = (φn − (1-φ)n ) ⁄ √5

This instantly leads to the approximation

Fn ≈ φn ⁄ √5

In fact, Fn is always the closest integer to φn ⁄ √5.

n Fn φn ⁄ √5
0 0 0.447214
1 1 0.723607
2 1 1.170820
3 2 1.894427
4 3 3.065248
5 5 4.959675
6 8 8.024922
7 13 12.984597
8 21 21.009519
9 34 33.994117
10 55 55.003636
11 89 88.997753
12 144 144.001389

Since 3 and 5 are the fourth and fifth Fibonacci numbers, we have that 3 is approximately φ4 ⁄ √5, and 5 is approximately φ5 ⁄ √5.

This means that 3*5 must be approximately φ9 ⁄ 5.

ie φ4 ⁄ √5 * φ5 ⁄ √5=φ9 ⁄ 5
3.06524758425* 4.95967477525 15.202631123499291
which implies 3* 5 15

So, the multiplicative identity 3*5=15 can be viewed as an integral approximation to the exact additive identity 4+5=9.

Mersenne primes

15 is 22*2-1.

There is an algebraic identity for nab - 1, namely

nab - 1 = (na-1)( 1+na+…+nab-a ).

The easiest way to see this is to realize that it's just
yb-1 = (y-1)(1+y+y2+…+yb-1), where y = na.

Substituting n=2, a=2, b=2, we get

22*2=(22-1)(1+22)

ie 15=(3)(5).

The above algebraic identity shows us that nm-1 cannot be prime if m is composite (given n ≥ 2 of course).

15 = 3*5 is the smallest example of the general fact "if q is composite, then 2q-1 is composite".

The above result about composite numbers is equivalent to the result "if 2p-1 is a prime, then p is also a prime".

Numbers of the form 2p-1 that are prime have been heavily studied, and have been given the name "Mersenne primes". Examples are 3 (= 22-1) 7 (= 23-1) and 31 (= 25-1). 15 = 24-1 can't be an example, since 4 is composite. Incidentally, p being prime doesn't necessarily imply that 2p-1 is prime. Consider 211-1 = 2047 = 23*89.

The largest known prime (as of August 2007) is the Mersenne prime 232,582,657-1

Product of Sum of Squares

Fibonacci showed that the product of the sum of two squares is always the sum of two squares. He did it by discovering the identity

(a2 + b2) (c2 + d2) = (ac-bd)2 + (ad+bc)2

Euler showed that the product of the sum of four squares is always the sum of four squares. He did it by discovering the identity

(a2 + b2 + c2 + d2) (e2 + f2 + g2 + h2) = (ae-bf-cg-dh)2 + (af+be+ch-dg)2 + (ag-bh+ce+df)2 + (ah+bg-cf+de)2

But, the product of the sum of three squares is NOT always the sum of three squares.

It turns out that the smallest example that demonstrates this fact about the sum of three squares is 3*5 =15

3 is the sum of three squares ( 12 + 1 2 + 12 )
5 is the sum of three squares ( 22 + 1 2 + 02 )
15 is not the sum of three squares.

Complex Numbers

Fibonacci's identity is really a fact about multiplication of complex numbers. The norm of a complex number z = a + bi is defined to be a2+b2 (equivalently (a+bi)*(a-bi)).

Writing z1 = a+bi and z2 = c+di, Fibonacci's identity is the statement

N(z1) N(z2) = N (z1z2)

Quaternions

The quaternions are a four dimensional generalization of the complex numbers.

For the quaternion a+bi+cj+dk, define the norm to be a2+b2+c2+d2 (equivalently (a+bi+cj+dk)*(a-bi-cj-dk)).

Writing q1 = a+bi+cj+dk and q2 = e+fi+gj+hk, Euler's identity is the statement

N(q1) N(q2) = N (q1q2)

If there were a three dimensional generalization of the complex numbers, then the corresponding norm construction would give us an identity for the product of the sums of three squares. The result "3*5=15" can be viewed as an obstruction to the existence of a three dimensional generalization of the complex numbers.

To be concrete, if we were to consider some generalization of the complex numbers that involved numbers of the form a+bi+cj, then 3*5=15 would tell us that there's going to be a problem with the product (1+i+j)*(2+i).

Of course, in the four dimensional world of quaternions, we have

(1+i+j)*(2+i) = 1+3i+2j-k

Applying the identity N(q1) N(q2) = N (q1q2), this becomes

(12+12+12)*(22+12) = (12+32+22+(-1)2)

3*5 = 15

An infinite product

So far, we've seen the identities:

x2-1 = (x-1)(x+1) (from the base 4 interpretation)

x4 - 1 = (x-1) (x+1) (x2+1) (from a base 2 interpretation)

This pattern can be continued, to yield further identities:

x8 - 1 = (x-1) (x+1) (x2+1) (x4+1)

x16 - 1 = (x-1) (x+1) (x2+1) (x4+1)(x8+1)

The x=10 versions are:
99 = 9*11
9999 = 9*11*101
99999999 = 9*11*101*10001
9999999999999999 = 9*11*101*10001*100000001

These algebraic identities can be rewritten as the sequence of identities:

1 - x = (1-x)

1 - x2 = (1-x) (1+x)

1 - x4 = (1-x) (1+x) (1+x2)

1 - x8 = (1-x) (1+x) (1+x2) (1+x4)

1 - x16 = (1-x) (1+x) (1+x2) (1+x4)(1+x8)

1 - x32 = (1-x) (1+x) (1+x2) (1+x4)(1+x8)(1+x16)

Taking the limit makes sense for small x (more precisely, for x with absolute value less than 1) and this leads to the infinite product:

1 = (1-x) (1+x) (1+x2) (1+x4)(1+x8)(1+x16)…

This tells us that:

1/1-x = (1+x) (1+x2) (1+x4)(1+x8)(1+x16)…

Multiplying out the infinite product, this becomes the familiar

1/1-x = 1 + x + x2 + x3 + x4 + x5 + …

Concluding words

I expect that there are many more mathematical results that can be somehow related to 3*5=15. Feel free to let me know about them.

Feedback

If you have corrections, additions, modifications, etc please let me know mailto:walterv@gbbservices.com

December 31 2003 Posted
August 8 2007 Last Updated

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