7 Things You Need to Know About Prime Numbers

Prime numbers are fundamentally important in mathematics. Watch this talk by Dr Vicky Neale (Mathematical Institute, University of Oxford) to discover some of the beautiful properties of prime numbers, and learn about some of the unsolved problems that mathematicians are working on today.

This talk was given to an audience of 16-17 year olds.

We'd love you to share how you've tried this and what you've discovered. You can add a comment below, or you can email us - your work may be featured in the showcase.

Comments

Thanks for the help!

Where can I get that pencil?

I've just asked Vicky. I'm afraid she was given the pencil and doesn't know where it came from.

She would be interested in having some more, so if anyone can find a supplier please let us know.

I thought that Vicky's talk was very interesting, and it made me think. If all primes are one more or one less than a multiple of 6, how can I prove this and why is it true?

Multiples of 6 have a digital root of 3, 6, or 9 (fact). The remaining digital roots are 1, 2, 4, 5, 7 and 8. Each of these are one more or one less than a multiple of 6 as a digital root, that is, one more or one less than 3, 6, or 9. Therefore, other than 2 and 3, all prime numbers must be one more or one less than a multiple of 6.

Vicky's pattern on her special maths pencil also shows a pattern in the digital roots (I like digital roots - a lot!!)

1st 2nd 3rd 4th 5th 6th
1 2 3 4 5 6
7 8 9 1 2 3
4 5 6 7 8 9
1 2 3 4 5 6
7 8 9 1 2 3
4 5 6 7 8 9

The third and sixth columns are multiples of 3, so are made up of 3, 6, and 9. All the other columns are just one more or one less than their neighbours.

Then I thought about normal remainders, rather than digital roots.

When you divide any number by 6, the remainder is 0, 1, 2, 3, 4, 5. If you add or subtract 0, 2 or 4 from any multiple of 6, you will end up with an even number: 6n ± 0 = 2(3n), 6n ± 2 = 2(3n ± 1), 6n ± 4 = 2(3n ± 2)
None of these can be prime, as they are all divisible by 2. Similarly, if you add or subtract 3 from any multiple of 6, you get: 6n ± 3 = 3(2n ± 1), which cannot be prime as it's divisible by 3.

We are only left with 1 and 5. If you add 1 to any multiple of 6, it is the same as subtracting 5 from the next multiple of 6:
6n + 1 = 6(n+1) - 5, and vice versa.

Therefore except for 2 and 3, all of the prime numbers can be expressed as 6n ± 1

One day I'd like to prove that there are infinitely many primes p, such that p+2 is also prime. I wondered whether remainders may be useful to play with, and this is what I imagined:

Any number divided by 3, can only leave 3 possible remainders: 0, 1, and 2. Adding 2 (twin prime difference) to each remainder gives (0 + 2 = 2), (1 + 2 = 3) and (2 + 2 = 4) Continue to simplify. I subtracted 3 when possible, to give a non-negative integer. This now gives 2, (3 - 3 = 0) and (4 - 3 = 1) It's important to understand that 0 creates 2, 1 creates 0 and 2 creates 1, because we are now going to eliminate options.

Remember, there were 3 remainders at the beginning: 0, 1 and 2, following division by 3. If any number leaves a remainder of 0, then it is divisible by 3 and cannot be prime, unless it is 3 itself. 0 creates 2, following addition of 2, so 3 and 5 are twin primes. Similarly, following the same process 1 creates 0, so cannot be prime (as it's divisible by 3) and 2 creates 1, which is the only possibility.

Since there are infinitely many numbers, there are infinitely many numbers that are divisible by 3, leaving infinitely many numbers which are not. Out of these, there are infinitely many that leave a remainder of 2, so there are infinitely many which will follow this pattern. I think this will also apply to every other prime number. I'd be interested to hear what you think of my ideas.

Finally, Vicky talked about two pairs of twin primes glued together where the middle prime is shared and gave the example of 3, 5, and 7 (three prime numbers, with a spacing of 2 between each). I don't think that there are any more like this because, every third number is a multiple of 3, so every set of three consecutive odd numbers must, by definition, contain a multiple of 3 (the difference of 2 matters - it leads to consecutive odd numbers). But 3 is the only multiple of 3 which is prime, therefore 3, 5 and 7 is the only set that satisfies Vicky's criteria.

Ta-dah!!

Just to be clear, looking at the pattern in the digital roots, all primes (except 3) have a digital root of 1, 7, 4, 5, 2, 8, because all primes are one more or one less than a multiple of 6; and 3 is the only prime with a digital root of 3.

The digital root pattern in the 1st column is repeated in the 4th column (but in a different order). Similarly, the digital root pattern in the 5th column is repeated in the 2nd column, again in a different order. In both cases, this is because of a constant difference of 3.

If I look at the numbers in the columns (rather than the digital roots) then column 2 contains numbers that are 2 plus a multiple of 6 (e.g. 0 x 6 = 0, 0 + 2 = 2), column 3 has odd multiples of 3, column 4 has numbers that are 4 plus a multiple of 6 (e.g. 0 x 6 = 0, 0 + 4 = 4) and column 6 contains multiples of 6.

I then thought about the primes 101,103 and 107,109. Both are twin primes. Both follow my rule: 101 divided by 3 leaves a remainder of 2, add 2 (twin prime difference) and simplify, i.e. subtract 3, I get a remainder of 1 (2 + 2 = 4, 4 - 3 = 1). And 103 divided by 3 leaves a remainder of 1 (fact). 107 and 109 follow the same pattern.

Other than 3, 5, 7, this set of quadruple primes are the closest that twin primes can be (there are others), because 101 + 6 =107, and 103 + 6 =109, again both follow my rule. The first number leaves a remainder of 2 after division by 3, and the second number leaves a remainder of 1 after division by 3 (because you are adding 6 and 6 is divisible by 3).

I'm going to think about this some more, because it's really interesting.

I wanted to think some more about this, because it's really interesting. Mum has explained modular arithmetic, and it's actually quite easy (I didn't realise I was already doing it).

My theory works for (mod 3), where the smaller (or lead) prime is congruent to 2(mod 3), and the larger (or second) prime is congruent to 1(mod 3), this is my twin prime theory for mod 3 which will hold for all twin primes except 3,5, as these are the only pair that follow 0(mod 3), 2(mod3). This is because all lead primes, in any twin prime pair (except 3), leave a remainder of 2 after division by 3 (fact).

I also know that it works for mod 5, look: in modulo 5, the only possible remainders are 0, 1, 2, 3, 4
Add 2 (the twin prime difference)

0(mod 5)+2=2≡ 2(mod 5)
1(mod 5)+2=3≡ 3(mod 5)
2(mod 5)+2=4≡ 4(mod 5)
3(mod 5)+2=5≡ 0(mod 5)
4(mod 5)+2=6≡ 1(mod 5)

So 0 creates 2, 1 creates 3, 2 creates 4, 3 creates 0 and 4 creates 1. Now to eliminate: 0(mod 5) is a multiple of 5, so cannot be prime, unless it is 5 itself, leading to the twin primes 5 and 7 which is the only pair possible in the form 0(mod 5), 2(mod 5). Similarly for 3(mod 5), 0(mod 5), the second number is a multiple of 5, so cannot be prime unless it is 5 itself, leading to the twin primes 3 and 5. Again, this is the only pair possible in the form of 3(mod 5), 0(mod 5).

The 3 other cases (1 creates 3, 2 creates 4 and 4 creates 1) all work for infinite pairs of twin primes. Some examples follow, but there will be many more:
11, 13 and 71, 73 both satisfy 1 creates 3, where 11≡ 1(mod 5), 13≡ 3(mod 5), and 71≡ 1(mod 5), 73≡ 3(mod 5);
17, 19 and 107, 109 both satisfy 2 creates 4, where 17≡ 2(mod 5), 19≡ 4(mod 5), and 107≡ 2(mod 5), 109≡ 4(mod 5);
29, 31 and 59, 61 both satisfy 4 creates 1, where 29≡ 4(mod 5), 31≡ 1(mod 5), and 59≡ 4(mod 5), 61≡ 1(mod 5); etc..

This will hold for all twin primes in modular 5 (except 3,5 and 5,7) because every lead prime in any twin prime pair ends in 1, 7 or 9 (except 3 and 5), and 1≡1(mod5), 7 ≡2(mod5), and 9 ≡4(mod5). I have already conjectured that my theory will work for every prime, but now I want to prove this.

But first I looked at modulo 2: where 0(mod 2)+2≡0(mod 2), 1(mod 2)+2≡1(mod 2). So 0 creates 0 and 1 creates 1.
0 creates 0 doesn't work, since if one of the numbers is 2, then the other cannot be prime as 2 is the only prime number that is a multiple of 2. But 1 creates 1 works in (mod 2), e.g. 41, 43 where 41≡ 1(mod 2), and 43≡ 1(mod 2).
Every pair of twin primes will follow the pattern 1(mod 2)+2≡1(mod 2), because they are all odd (fact).

Proof: let the lead prime be p, and it's twin prime be p+2. Since all twin primes are odd, let p = 2n+1

(2n+1) = 1+2n≡1(mod2)
(2n+1)+2 = 3+2n= 1+2(n+1)≡1(mod2)

Now for a general twin prime solution:

Because I am looking for options that generate infinite possibilities, single cases don't matter and are eliminated.

Consider (mod p), where p is a prime number ≥3.

In every case, I would eliminate 2 options: (a) where the first number is 0(mod p) and (b) where the second number is 0(mod p), because both are divisible by p, and therefore composite, unless p itself is the first or second prime in a twin prime pair.

Because I am looking for infinite possibilities, NOT single cases, where p itself is the first or second prime in a twin prime pair, it only works once, e.g. 5≡0(mod5)+2=7≡2(mod5), or 3(mod5)+2=5≡0(mod5), etc...It doesn't work for multiple primes, so I will eliminate these options. This will leave at least 1 possibility for (mod p) e.g. if p = 3.

So in (mod p) there are p possibilities, but 2 of them will not work for multiple primes, leaving at least p-2 possibilities that do. The smallest value for p which leaves a positive integer is 3, as 3-2=1, and 2 is the twin prime difference. That is why I started with 3 in my earlier work, but now I can explain it using modular arithmetic.

In modulo 3, there are 3 possibilities; 0, 1, 2; however 2 of the possibilities don't work for multiple pairs of twin primes. 0 creates 2 only works for the twin prime pair 3, 5 (a single case); and 1 creates 0 doesn't work in any circumstance, as the second number is composite (a multiple of 3, following addition of 2). The only possibility that generates multiple twin prime pairs in modulo 3 is 2 creates 1, or 2(mod 3)+2≡1(mod 3). This is because all lead primes, in any twin prime pair (except 3), leave a remainder of 2 after division by 3 (fact).

To make it clearer, in modulo p, following division by p, there are a total of p remainders: 0, 2, 3, 4, ..... p-1. Two of these will not work for multiple primes; as they generate a number that is congruent to is 0(mod p), leading to a single case or a composite number. This leaves at least p-2 possibilities for multiple twin prime pairs.

Since there are infinitely many positive integers, there are infinitely many that are divisible by p, leaving infinitely many which are not. Out of these, there are infinitely many that don't leave a remainder of 0 or p-2, so there are infinitely many which will follow this pattern. Therefore I can conclude that there are infinitely many twin primes.

This is my conjecture for twin primes, totally inspired by Vicky's talk on primes.

To test my theory, I chose a random prime number:
Let p=53. The possible remainders after division by 53 are 0, 1, 2, 3, ...... 52.
0(mod 53)+2≡2(mod 53) and 51(mod 53)+2≡0(mod 53) (Note that p-2=51)
0(mod 53) is a multiple of 53 so cannot be prime unless it is 53 itself, but (53-2=51) and (53+2=55) are both composite. So if p is 53, there are only p-2 or 51 possibilities to generate multiple twin primes. Again, working through the numbers will make it clearer. I have only listed one pair of twin primes in each case, but there will be more:

0(mod 53) as I've explained, this creates a composite number, so doesn't work.
1(mod 53) works, e.g. 107, 109 where 107≡1(mod 53) and 109≡3(mod 53)
2(mod 53) works, e.g. 4931, 4933 where 4931≡2(mod 53) and 4933≡4(mod 53)
3(mod 53) works, e.g. 1487, 1489 where 1487≡3(mod 53) and 1489≡5(mod 53)
and this continues, so similarly:
4(mod 53) works, e.g. 269, 271
5(mod 53) works, e.g. 641, 643
6(mod 53) works, e.g. 59, 61
7(mod 53) works, e.g. 431, 433
8(mod 53) works, e.g. 2711, 2713
9(mod 53) works, e.g. 857, 859
10(mod 53) works, e.g. 1229, 1231
11(mod 53) works, e.g. 2237, 2239
12(mod 53) works, e.g. 1019, 1021
13(mod 53) works, e.g. 2027, 2029
14(mod 53) works, e.g. 809, 811
15(mod 53) works, e.g. 227, 229
16(mod 53) works, e.g. 281, 283
17(mod 53) works, e.g. 1289, 1291
18(mod 53) works, e.g. 71, 73
19(mod 53) works, e.g. 4259, 4261
20(mod 53) works, e.g. 179, 181
21(mod 53) works, e.g. 2141, 2143
22(mod 53) works, e.g. 1877, 1879
23(mod 53) works, e.g. 659, 661
24(mod 53) works, e.g. 1031, 1033
25(mod 53) works, e.g. 1721, 1723
26(mod 53) works, e.g. 821, 823
27(mod 53) works, e.g. 239, 241
28(mod 53) works, e.g. 12377, 12379
29(mod 53) works, e.g. 347, 349
30(mod 53) works, e.g. 2309, 2311
31(mod 53) works, e.g. 137, 139
32(mod 53) works, e.g. 191, 193
33(mod 53) works, e.g. 881, 883
34(mod 53) works, e.g. 617, 619
35(mod 53) works, e.g. 3851, 3853
36(mod 53) works, e.g. 1997, 1999
37(mod 53) works, e.g. 461, 463
38(mod 53) works, e.g. 197, 199
39(mod 53) works, e.g. 569, 571
40(mod 53) works, e.g. 3167, 3169
41(mod 53) works, e.g. 1949, 1951
42(mod 53) works, e.g. 1049, 1051
43(mod 53) works, e.g. 149, 151
44(mod 53) works, e.g. 521, 523
45(mod 53) works, e.g. 2801, 2803
46(mod 53) works, e.g. 311, 313
47(mod 53) works, e.g. 1319, 1321
48(mod 53) works, e.g. 101, 103
49(mod 53) works, e.g. 1427, 1429
50(mod 53) works, e.g. 1481, 1483
51(mod 53) as I've explained, this creates a composite number, so doesn't work.
52(mod 53) works, e.g. 6359, 6361

It took me ages to find 28(mod 53), but I did find it eventually - truly unprecedented!!

So what does this tell me? I think there's a pattern here... so I went back to Vicky's special pencil. The twin primes I had found were the first ones that worked in each circumstance (there will be more). I arranged these over six columns. But instead of looking at the lead prime number, I calculated the multiple of 53 it represented and worked out its digital root. This is what I found:

Columns are labelled in bold: (X = composite) and shows the actual multiples of 53:

1 2 3 4 5 6
X 2 93 28 5 12
1 8 51 16 23 42
19 38 15 4 5 24
1 80 3 40 35 12
19 32 15 4 233 6
43 2 3 16 11 72
37 8 3 10 59 36
19 2 9 52 5 24
1 26 27 X 119

Then I noticed something:
Column 6 = multiples of 6
Column 5 = multiples of 6, minus 1
Column 4 = multiples of 6 minus 2
Column 3 = multiples of 6 minus 3
Column 2 = multiples of 6 minus 4
Column 1 = multiples of 6 minus 5
This means I can rewrite these numbers in mod 6:

1, 19, 43, 37 are ALL congruent to 1(mod 6)
2, 8, 38, 80, 32, 26 are ALL congruent to 2(mod 6)
93, 51, 15, 3, 9, 27 are ALL congruent to 3(mod6)
28, 16, 4, 40, 10, 52 are ALL congruent to 4(mod 6)
5, 23, 35, 233, 11, 59, 119 are ALL congruent to 5(mod 6)
12, 42, 24, 6, 72, 36 are ALL congruent to 0(mod 6)

I then calculated the digital roots of these numbers:

1 2 3 4 5 6
X 2 3 1 5 3
1 8 6 7 5 6
1 2 6 4 5 6
1 8 3 4 8 3
1 5 6 4 8 6
7 2 3 7 2 9
1 8 3 1 5 9
1 2 9 7 5 6
1 8 9 X 2

Remember, I am no longer looking at remainders, but the actual multiples of p in modulo p, expressed as digital roots.

In my second piece of work, I stated that all primes (except 3) have a digital root of 1, 7, 4, 5, 2, 8.

In my first piece of work I explored Vicky's pattern for primes and found that the digital roots in columns 1 and 4 were {1,4,7}, in columns 2 and 5 were {5,2,8} and in column 3 and 6 were {3,6,9}. The same pattern appears in my chart. Remember that I am NOT looking at primes here, but rather integer multiples of p in modulo p. These are just numbers, not actual primes. Although a 4 doesn't appear in the 1st column, I predict that this would, if I started with a different value for p.

So does this mean that I can predict where twin primes appear? It's an interesting question, and one that I will continue to think about. But I still wanted to know why this works. So I thought about digital roots (DR).

Take for example, 8(mod 53) generates the lead prime 2711 (in 2711, 2713).
2711-8=2703÷53=51 these are the numbers,
DR of 2711=2, DR of 8=8, DR of 2703=3, DR of 53=8, DR of 51=6

The corresponding sum using digital roots is:
2-8 is rephrased as 11-8=3 leading to 3÷8, rephrase again to 48÷8=6 which is why it works!

I really did enjoy doing this, it is beautiful when all the numbers fall into place and although I'm only just 10, I will continue to think about this so that I can solve it when I grow up. I've made new discoveries and learnt to express myself in modular arithmetic. Thank you Vicky, your talk was brilliant!!

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