Is a Pattern That Works Ten Times Always True?

The pattern breaks

If you counted carefully, your results read 1, 2, 4, 8, 16 and then 31. One piece short. Try 7 points and the gap grows: doubling says 64, the circle gives 57. With 8 points it is 99 pieces, not 128. The pattern didn’t just slip by one; it was never the rule at all.

Points	1	2	3	4	5	6	7	8
Pieces	1	2	4	8	16	31	57	99
Doubling	1	2	4	8	16	32	64	128

Five correct cases in a row felt like proof. They weren’t. This is the whole topic in one picture: a pattern that works many times can still be false, and the only way to be sure is to find out why it works.

The meeting point

There is a catch in the puzzle, and the widget shows it. Press Spread evenly and try 6 points again. Now there are only 30 pieces: the three longest lines all pass through the very center, and where three lines meet at one point, a little piece that would have been there is squeezed away to nothing. Try to picture it: nudge one dot a tiny bit, and the three lines no longer meet at one point; a tiny triangle opens up in the middle.

So the question needs care. The most pieces you can get with 6 points is 31, and you get it whenever no three lines meet at one point. That is the question mathematicians ask, and it is the one we answer below. Spread evenly, 8 points make 88, not 99.

Guess, test, prove

Sushila and Ramesh did what every mathematician does. They looked at some cases and made a guess: “the pieces double”. A guess like that, believed but not yet shown to be true, is a conjecture. Then they tested it, and the test found a case where it fails, a counterexample. One counterexample is enough to kill a conjecture, however many cases fit.

To make a conjecture into a fact you need a proof: an argument that shows why it is true, so that it must hold in every case, including the ones nobody has drawn. So let’s find out why the circle gives the numbers it does. Then we will also see why it pretended to double.

Each line adds 1, and 1 more for every crossing

Draw the lines one at a time and watch the pieces grow. Before any line, the circle is 1 piece.

Now draw a new line. It runs from one edge of the circle to the other, and on the way it crosses some of the lines already there. In Figure 16.1 the blue line crosses 2 of them, and the 2 crossings cut it into 3 bits. Each bit runs straight across one piece and splits it in two. So the blue line adds 3 pieces: one more than the number of lines it crosses.

That is true for every line. A line that crosses nothing (the first one, say) adds 1 piece. A line that crosses 4 others is cut into 5 bits and adds 5. So when every line is drawn:

pieces = 1 + (the number of lines) + (the number of crossings inside the circle)

because each line gives its 1, and each crossing gives 1 more, once, when the second of its two lines is drawn. (This is where “no three lines through one point” matters. Three lines through one point make three pairs crossing at the same spot, but only one point there to cut at, so you lose pieces. That is the 30.)

Counting lines and crossings

The lines. Every pair of points is joined by one line, just as every pair of guests at a party shakes hands once. In How Many Handshakes at a Party of 30? you found that n people make ${}^{n}C_{2}$ handshakes, which is half of n × (n − 1). So 6 points make 6 × 5 ÷ 2 = 15 lines.

The crossings. Here is the clever part. Take any 4 of the points, like A, B, C and D in Figure 16.2, and draw all six lines between them. The four lines around the outside never cross each other inside the circle. Only the two lines that join opposite points, A to C and B to D, cross, and they cross exactly once. So every group of 4 points gives exactly one crossing, and every crossing comes from exactly one group of 4 points: the ends of its two lines.

So the number of crossings is the number of ways to choose 4 of the points, ${}^{n}C_{4}$. For 6 points there is a neat way to count them: choosing 4 of the 6 to keep is the same as choosing 2 of the 6 to leave out, and that is 15 again. For any number of points, pick the 4 one at a time and divide by the 4 × 3 × 2 × 1 = 24 orders the same group can be picked in, as in How Many Ways Can You Walk to School?: 6 points give 6 × 5 × 4 × 3 ÷ 24 = 15.

Now the 6 points add up: 1 + 15 lines + 15 crossings = 31. Proved, not guessed, and it works for any number of points:

Points	Lines	Crossings	Pieces = 1 + lines + crossings
4	6	1	8
5	10	5	16
6	15	15	31
7	21	35	57
8	28	70	99

Why it looked like doubling

So why did 1, 2, 4, 8, 16 come out so neatly? The answer is hiding in Pascal’s triangle, which you can meet in How Many Ways Can You Walk to School?. Each row of the triangle adds up to double the row before: 1, 2, 4, 8, 16, 32… That is where doubling lives.

Now look at Figure 16.3. For each number of points, take its row (for 6 points, 1, 5, 10, 10, 5, 1) and add up only its first five numbers. You get exactly the circle’s pieces, every time: 1 + 5 + 10 + 10 + 5 = 31.

It works for every row, and the triangle’s own rule, each number the sum of the two above it, shows why. The 15 lines and 15 crossings of 6 points are the two 15s in the row 1, 6, 15, 20, 15, 6, 1; from the row above, the first is 5 + 10 and the second 10 + 5. So 1 + 15 + 15 = 1 + 5 + 10 + 10 + 5, the first five numbers of 1, 5, 10, 10, 5, 1, and the same happens in every row. (John Conway and Richard Guy found another proof, matching each piece with a group of at most four of the points.)

Up to 5 points, the rows are short: they have five numbers or fewer, so “the first five” is the whole row, and the pieces really do double. At 6 points the row has six numbers, and the last one, 1, is left out: 32 − 1 = 31. After that, more and more is left out each time: 64 − 7 = 57, 128 − 29 = 99. So the doubling was never the rule. It is what the real rule happens to give while the rows are short, and the first five cases were all too small to show the difference.

The circle is not the only pattern that fools people. The widget has three more that did, and one that never breaks. Test each one yourself before you read on.

Forty primes in a row

You met the first in Is There a Biggest Prime Number?: Leonhard Euler’s formula n² + n + 41 (n × n + n + 41), noticed in 1772. It gives a prime for every n from 0 to 39, forty in a row, and then at n = 40 it gives 1,681 = 41 × 41. Asking why it might fail finds a counterexample without any testing: at n = 41 every part has 41 in it, 41 × 41 + 41 + 41 = 41 × 43.

31, 331, 3331…

31 is prime. So is 331. And 3,331, and 33,331, and 333,331, and 3,333,331, and 33,333,331: seven primes in a row, made by the same rule (put another 3 in front). It is very hard to believe that the eighth won’t be prime too. But 333,333,331 = 17 × 19,607,843. The puzzle writer Martin Gardner called it “a remarkable pattern that is entirely accidental and leads nowhere”. Unlike Euler’s formula, there is no reason it should have held even seven times: it just did.

Goldbach’s other guess

The third one fooled a great mathematician. Christian Goldbach, in a letter to Euler, guessed that every odd number that isn’t prime is a prime plus twice a square number: 9 = 7 + 2 × 1², 15 = 13 + 2 × 1², 27 = 19 + 2 × 2², 33 = 31 + 2 × 1².

Test it in the widget. It works for 9, for 15, for 21, for 25… for 2,131 odd numbers in a row (every odd number from 9 to 5,775 that isn’t prime). Then comes 5,777 = 53 × 109. Subtract twice every square smaller than it, 2, 8, 18, 32, … up to 2 × 53², and not one of the answers is a prime. The guess is false. Odd numbers that can’t be made this way are named after the German mathematician Moritz Stern, who studied them. Only two of them aren’t prime: 5,777 and 5,993 = 13 × 461. Nobody has found a third, and computers have checked every odd number up to 2 × 10¹³.

The one that never breaks

The fourth pattern in the widget is different: 1 = 1², 1 + 3 = 4 = 2², 1 + 3 + 5 = 9 = 3², 1 + 3 + 5 + 7 = 16 = 4². Adding up odd numbers from 1 always gives a square. Press Keep testing as often as you like: it never fails. But by now you know that passing 10,000 tests, or a million, proves nothing. So is it true? Here is a proof.

Look at Figure 16.4. Start with 1 tile: a 1 × 1 square. Wrap an L of 3 tiles around it and you have a 2 × 2 square. Wrap an L of 5 around that: a 3 × 3 square. An L of 7, then 9: 4 × 4, then 5 × 5.

Why does it go on forever? Take any square of tiles, say 10 × 10. To wrap an L around it you need a row of 10 along the bottom, a column of 10 up the side, and 1 tile for the corner: 10 + 10 + 1 = 21 tiles, the next odd number after 19. And 10 × 10 with an L of 21 is 11 × 11. That works for any size, not only 10: the next L is always the next odd number, and it always makes the next square.

So the pattern is true for 1 tile, and whenever it is true for one square, it is true for the next. Like a row of dominoes where the first one falls and each one knocks over the next, it must be true for every square, all the way up. This way of proving is called proof by induction, and it is how mathematicians prove that a pattern holds for every whole number. It needs two things: a first case that is true, and a step that carries the truth from each case to the next. Testing gives you the first; only a reason gives you the step.

Checked a billion billion times, still not proved

Some patterns are tested by computers far beyond anything a person could check, and still nobody knows whether they are true. The most famous is another of Goldbach’s guesses, from his letters with Euler in 1742: every even number bigger than 2 is the sum of two primes (4 = 2 + 2, 10 = 3 + 7, 100 = 47 + 53). It has been checked for every even number up to 4 × 10¹⁸, four billion billion, by a computer search led by Tomás Oliveira e Silva (published in 2014). Nobody has found a counterexample. Nobody has found a proof either. It is still a conjecture, called Goldbach’s conjecture, and as of 2026 it is one of the most famous unsolved problems in mathematics.

After everything above, you can see why mathematicians won’t call it a fact. A counterexample could be bigger than any number checked so far.

When the counterexample is enormous

How big? Here is one that waited a long time. Write a number as primes multiplied together, counting repeats: 18 = 2 × 3 × 3 has 3 primes, 60 = 2 × 2 × 3 × 5 has 4. In 1919 the Hungarian mathematician George Pólya guessed that, from 1 up to any number you like beyond 1, at least half of the numbers have an odd number of primes (1 itself counts as having none). It holds up to 10, up to 1,000, up to a million, up to a hundred million. The first number where it fails is 906,150,257, found by Minoru Tanaka in 1980.

• Where Did the Missing Square Go? A picture that seems to prove something impossible, and the flaw hiding in it: a proof can fool you as well as a pattern.
• Who Is Telling the Truth? Proof by cases and proof by contradiction, the other two ways of being sure, on an island of truth-tellers and liars.
• Can You Prove √2 Is Not a Fraction? (Grades 9–12) A famous proof by contradiction.
• How Many Moves to Move 64 Disks? (Computing, Going Further) The Tower of Hanoi: its fewest moves follow a pattern you can prove by the domino idea, and program.
• Will This Simple Rule Ever Stop? (Computing, Going Further, Grades 9–12) Halve it if it’s even, triple it and add one if it’s odd: a pattern checked by computers for huge numbers, and still not proved. (Coming soon.)
• Grade 10, Sequence and Series: patterns of numbers with rules you can prove, and their sums.