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7 is prime so only factorable by 1 or 7.

the only squares less than 6 are 1 and 4 therefore a=1, r=2

a=7, r=0

73 is prime so only factorable by 1 or 73. 73 too big so must be 1. Only smaller than 73 is r= 1 or 2. 1 is too small so a=1, r=2.

Examples: [Formulas followed by prime factorization, as you can see factors 3,7,11,19,23,31 and 43 are missing]

n=1

n=2 5

n=3 10=2,5

n=4 17

n=5 26=2,13

n=6 37

n=7 50=2,5

n=8 65=5,13

n=9 82=2,41

n=10 101

n=11 122=2,61

n=12 145=5,29

n=13 170=2,5,17

n=14 197

n=15 226=2,113

n=16 257

n=17 290=2,5,29

n=18 325=5,13

n=19 362=2,181

n=20 401

n=21 442=2,13,17

n=22 485=5,97

n=23 530=2,5,53

n=24 577

n=25 626=2,313

n=26 677

n=27 730=2,5,73

n=28 785=5,157

n=29 842=2,421

n=30 901=17,53

n=31 962=2,13,37

n=32 1025=5,41

n=33 1090=2,5,109

n=34 1157=13,89

n=35 1226=2,613

n=36 1297

n=37 1370=2,5,137

n=38 1445=5,17

n=39 1522=2,761

n=40 1601

n=41 1682=2,29

n=42 1765=5,353

n=43 1850=2,5,37

**Primenumbers**- Replies: 2

New Theorem:

Potential factors for

Because.............

Let p= potential factor

Let p=n-y

(n-y)(n+y) =

Therefore remainder when is divided by p =

And remainder when is divided by p =

y will be less than p because y=n-p and p>0

p will not be greater than n because we are only concerned with primes

This should reduce potential factors for

and reduce computing space for finding primes.I have tried up to n=43. Prime factors <44 that were missing were 3, 7, 11, 19, 23, 31 and 43 which exactly halved the no. of potential prime factors. Any value for n to infinity will not be factorable by these no.'s.

The factors repeat themselves every +p value for n. So if they don't occur before

**Primenumbers**- Replies: 3

Fermat's Last Theorem states that no three positive integers a, b, and c satisfy the equation

for any integer value of n strictly greater than two.Let

where n >2Then

must equal:must have opposite remainders for

Adding y to one and minusing y from another will keep them opposite if they are already opposite. So let's see if the two squares have opposite remainders for .

If we add them together they should=0 if they are opposite.

No!

Fermat's Last theorem is True.

I don't know, it's just a Maths degree, starting 2017. Doing Foundation Year and Placement Year.

Maths now.

I got into Uni to do maths! I am going to do that now.

Q: Why did the angry Jedi cross the road?

A: To get to the Dark Side.

No way! That's not awesome.

Thirty and counting!

Can be written as....

And we know

So,

569,673=(1x2)+(2x4)+(4x6)+(6x8)+(10x12)+(12x14)+(16x18).

Well usually you'd do

where x=(2-1)(3-1)(5-1)(7-1) and it would be factorable by 2,3,5,7........But if you use a composite Ie. it won't be factorable by 3 and 7.

In this way you can find out if the base is prime or not in one simple calculation!

Oh! Fair enough.

**Primenumbers**- Replies: 8

**Introduction**

The following is a way to find out if (a) is prime or not using Fermat's Little Theorem.

**Rule**

x=(All primes below square root a)-1, multiplied together. I.e. (2-1)(3-1)(5-1)(7-1)(11-1)....

a=any odd no.

**Theory**

Fermat's little theorem states that if p is a prime number, then for any integer a, the number

is an integer multiple of p.So if y=p-1 then will be factorable by p.

We know that remainders repeat themselves in a pattern when multiplying by a. So y could be a multiple of y and the answer would still be factorable by p.

In this way as long as x is factorable by p-1 then the answer will be factorable by p unless (a) itself is factorable by p.

**Example**

a=37

x=(2-1)(3-1)(5-1)=8

a=21

x=(2-1)(3-1)=2

** Conclusion**

I don't know but I would have thought this method would be useful for computers as it only requires a few computations. Don't computers usually have to factor possible factors separately? The numbers are very large though.

Okay bobbym. I am going to bed now. 6 o'clock in the morning where I am.

Night night.

Okay, that's fair.

My avatar is because I used to play judo.

I tend to do everything to the extreme also. My friends used to call me Xtreme when I was a bit younger because I used to do everything to the extreme. Jump off things, drink the most and drive stupidly, stuff like that. I have calmed down a lot now but still enjoy a good challenge! I am at a weird point in my life when I am finally feeling better and I just don't know what to do with myself.

Mmm....do many people die from sailing round the world?

bobbym wrote:

That she is a she.

What made you think I was a guy bobbym?

I have done some research in case anyone else is interested in climbing it.

It's going to set you back something like £40, 000.

You need to;

Go on a climbing course and then practice

Go on an ice climbing course and then climb a lesser mountain

Do a load of training before attempting the ascent

Also you should know;

70% of attempts are unsuccessful and for every 10 climbers that make the ascent 1 will die trying to

That is a high risk! I'm quite scared now..................................................... Oh well!