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The law of small numbers

There is order in the Universe:

Most people will see some kind of pattern in the figures above. Pause a minute and write down on a piece of paper what you see.

Will the patterns continue? A good question. Can you prove that they will continue? An even better question.

There are many published examples of patterns that continue forever, that do not break down after a while. Could you imagine that 'The sum of the first n odd numbers equals n² except when n = 23,982?' I can't imagine such a theorem. I can't imagine that such a simple system can break down. I feel don't need a proof.

The urge to leap from a few cases to an infinite number is as psychologically strong as it is logically wrong. A few swallows do not make a mathematical summer, although we'd love to think so.

As part of our mathematical diet we need to experience simple patterns that do break down. There is a difference between feelings and truth, between what our eyes see and what is there, between imagination and reality.

Take a reality check.

Answer these questions to test your intuition, your feeling for things:

1. A pancake. Mark some points around the edge. Make a straight cut between all pair of points. If you start with two points you get the pancake divided in two pieces. With three points you get four pieces. With four points you get eight pieces. The question is: will the doubling continue? Will you be able to get sixteen pieces with five points, thirty-two with six points, and so on?

2. Choose a pair of natural numbers x and y. Do they fit in the equation x² - 991y² = 1? If not, try another pair. Try many. Can you find a solution? Do you think one exist?

If you have two wrong out of two so far, I throw in a few easy ones.

3. 3 is a prime number, 5 is a prime, 7 is a prime. "By George, I believe every odd number is prime!" Do you agree?

4. How many prime numbers exist? Let's make a table:

There seems to be fewer and fewer prime numbers as we move along. I guess by 100 there will be noone left. How many prime numbers do you think there are?

5. 2ⁿ will never be a prime number, regardless which natural number we use for n. But what about the  2ⁿ + 1. Maybe this number is always a prime?

No luck. 9 is not a prime. Let's have another shot. What about ?

n	0	1	2	3	4
	3	5	17	257	65,537

All of these are primes! Do you think is always a prime number?

6. Lets play a bit with the even numbers 2, 4, 6, 8, ...  Add 41 to 2 and you get 43. Add 43 to 4 and you get 47. Add 47 to 6 and you get 53. Add 53 to 8 and you get 61. It so happens that 41, 43, 47, 53 and 61 are all prime numbers. Will this pattern continue for ever?

7.The statement "n < 1,000,000,000" is true for all n I have tested so far. Will it ever break down? (Relax! It is just a joke!)

8. Look at this pattern:

Let me come with a modest claim. The division in this pattern will always give a whole number. Agree?

9. Pick a number p. Can you find a number n > 1such that the sum of the digits of n^p equals n? Let's look at p = 1, 2 and 3:

n^p -> the sum of the digits of n^p
2¹ = 2 -> 2
9² = 81 -> 9
8³ = 512 -> 8
will it continue?

10. "Every odd number is the sum of a prime and a power". Example: 15 = 11 + 2². True or false for all odd numbers?

11. "Every odd number is the sum of a prime and twice a square." Example: 31 = 23 + 2 * 2². True or false for all odd numbers?

12. What's the remainder of  2^x ÷ x? Let's have a look:

In the table the remainder is 0, 1, 2 and 4. My bold guess: The remainder is never equal to 3. Agree for all x?

13. "n⁵ + 5 and (n+1)⁵ + 5 has no factor in common." Agree for all n?

14. Look at this pattern and then guess the answers to the last two divisions to one decimal place. (From Mathematics Teaching no 70, page 35.)

987654321 ÷ 123456789 = 8.0
87654321 ÷ 12345678 = 7.1
7654321 ÷ 1234567 = 6.2
654321 ÷ 123456 = 5.3
54321 ÷ 12345 = 4.4
4321 ÷ 1234 = 3.5
321 ÷ 123 = 2.6
21 ÷ 12 = ? 
1 ÷ 1 = ? 

15. In January 1999 this message was posted on a newsgroup:

Hi,
Why are there no square numbers (apart from 1 and 9) that are made up completely of odd digits? I don't know this to be true but I've tested them all up to about 36,000,000,000,000 so I'm guessing it is.

What would you answer?

Solutions:

1. 16 can be done, but 32 is impossible! 31 is the most you can make with six points. For more information, see Ogilvy, C Stanley. Tomorrow's Math. Unsolved Problems for the Amateur. Oxford UP 1972.

2. x² - 991y² = 1 can be solved with x and y integeres.
The smallest solution is x = 379,516,400,906,811,930,638,014,896,080 and
y = 12,055,735,790,331,359,447,442,538,767.

Here is a check, using Maple V:

3. 9 is not a prime.

4. There are infinitely many primes.

5. Pierre de Fermat (1601-1665) wondered if would always give a prime. In 1729, 64 years after Fermat's death, Leonard Euler showed that it was composite for n = 5. In 1880 it was factorised for n = 6. Now the guess is "There are no more primes of this form than those five at the beginning." No one knows if this is true or not.

n	0	1	2	3	4	5	6
	3	5	17	257	65,537	641 * 6,700,417	274,177 * 67,280,421,310,721

6. The numbers are on the form n² - n + 41 with n = 0, 1, 2, ... When n = 41 we get 41², obviously not a prime.

7. (The joke.)

8. It is false, but I don't know when it breaks down. (If you do, tell Mum.) Please see below for an email from a reader.

9. This is true all the way up to p = 104, and fails at p = 105.

10. 1,549 is the smallest odd number not the sum of a prime and a power.

11. 5,777 is the smallest odd number not the sum of a prime and twice a square.

12. Smallest counter example is for x = 4,700,063,497.
2^{4,700,063,497} ÷ 4,700,063,497 has remainder 3.
By the way, 2^{4,700,063,497} has about 10¹⁰ digits, so there is no room in the margin...

13. The smallest counter example is for n = 533,360.
533,360⁵ + 5 = 5 * 1968751 * 55641478729429717 * 78803 and
533,361⁵ + 5 = 2 * 31 * 1968751 * 549756587 * 643210846049. The common factor is in bold.

14.
21 ÷ 12 = 1.8 
1 ÷ 1 = 1.0

15. This is one answer that was given in the group:

And ... you did not see a pattern in the possibilities for the last two digits?

Oh, I dunno... I mean, 36 million million is a pretty small number
(compared to infinity)... maybe he needs to try a few more just to
be sure.... ;-)

References:

• Guy, Richard K. The strong law of small numbers. (English) American Mathematical Monthly 95 (1988), no. 8, 697-712.
• Ogilvy, C Stanley. Tomorrow's Math. Unsolved Problems for the Amateur. Oxford UP 1972.
• Wells, David. The Penguin Dictionary of Curious and Interesting Numbers. Penguin 1988.

Reactions:

This space is for your reactions to the thoughts above. Send an e-mail to Mum.

Email received from Aleksi Vähäkangas 26th of June 2004.

Hi!

I just visited your page "The law of small numbers" and
this concerns the problem 8 there.

I think that the first element that is not a whole
number is the 43rd:

Denote
  x_1=1+1,
  x_2=1+1+(x_1/1)^2,
  x_3=1+1+(x_1/1)^2+(x_2/2)^2,
  ...
  x_{i+1}=x_i+(x_i/i)^2.

We are interested in the smallest n s.t. x_n/n is not
an integer, that is: smallest n s.t. x_{n+1} is not
integer.

Suppose that x_i is integer for all 1<=i<=n, n>4
and suppose that N>0 is divisable by n!.

Define
  y_1=1+1,
  y_2=1+1+(y_1/1)^2  mod N.
Then x_1=y_1 and
  y_2=x_2 mod N
In particular y_2 is divisable by 2 (since x_2 is) and
  y_2/2=x_2/2 mod N/2,
hence
  (y_2/2)^2=(x_2/2)^2 mod N/2.

Define integer
  y_3=y_2+(y_2/2)^2  mod N.
Then
  y_3=x_3 mod N/2.
In particular y_3 is divisable by 3 (since x_3 is) and
  y_3/3=x_3/3 mod N/(2*3),
hence
  (y_3/3)^2=(x_3/3)^2 mod N/(2*3).

Define integer
  y_4=y_3+(y_3/3)^2  mod N.
Then
  y_4=x_4 mod N/(2*3).
In particular y_4 is divisable by 4 (since x_4 is)...

......

Suppose that 3<=i<=n-1 and that
(*)  y_i=x_i mod N/(2*3*4*...*(i-1)).

Then, in particular y_i is divisable by i (since x_i is) and
     y_i/i=x_i/i mod N/(2*3*4*...*i),
hence
(**) (y_i/i)^2=(x_i/i)^2 mod N/(2*3*4*...*i).
Define integer
     y_{i+1}=y_i+(y_i/i)^2  mod N.
Then by (*) and (**) we have
     y_{i+1}=x_{i+1}  mod N/(2*3*4*...*i).

By induction we see that
  y_n=x_n mod N/(2*3*4*...*(n-1)).
We assumed that N is divisable by n! and hence
  y_n=x_n mod n.
In particular x_n/n is integer iff
  y_n=0 mod n.

The benefit of going from x_i to y_i is that y_i are quite small
compared to x_i. The following maple code
computes y_i for N=n!, where n=43:

>n:=43;N:=n!:

                               n := 43

>y[1]:=1+1:
>for i from 1 to n-1 do   y[i+1]:=(y[i]+(y[i]/i)^2) mod N; end do:

>for i from 1 to n do   [i,y[i] mod i];
>end do;
>
                                [1, 0]
                                [2, 0]
                                 ...
                               [40, 0]
                               [41, 0]
                               [42, 0]
                               [43, 24]

This shows that
  (i) x_i/i is integer for all 1<=i<=42,
  (ii) x_{43}/43 is not integer, in fact
       x_{43}=24  mod 43.

If you think I have made some mistake, feel free to correct me.

-Aleksi Vähäkangas

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