## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Math - Prime number

87 replies to this topic

### #81 Apolo Program

Apolo Program

Newbie

• Members
• 7 posts

Posted 16 February 2009 - 08:30 PM

Spoiler for Solved!!

• 0

### #82 Udontn0me

Udontn0me

Newbie

• Members
• 1 posts

Posted 26 February 2009 - 02:37 AM

just to let everyone know i didnt read pages 2-8 so if this was put up then sry
Spoiler for mathematical ingenius

• 0

### #83 paperphoenix

paperphoenix

Newbie

• Members
• 12 posts

Posted 26 February 2009 - 03:59 AM

Wow, so may formulas...

Edited by paperphoenix, 26 February 2009 - 04:01 AM.

• 0

### #84 Ryan91

Ryan91

Newbie

• Members
• 1 posts

Posted 23 March 2009 - 03:20 PM

wow...if it's [a*a] + 26 then i would say [2*2] + 26 = 30 but idk if i'm wrong or not...lol im not a mathematician (or a good speller) but i kno basics...and multiplication comes first so therefore it would eventually be 4+26=30...am i right?
• 0

### #85 PGupta

PGupta

Newbie

• Members
• 4 posts

Posted 26 June 2009 - 07:50 PM

This is a long explanation. I tried to be thorough. Let me know if you find any flaws.

Consider the set of perfect squares:
{ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ... }

0+1 = 1
1+3 = 4
4+5 = 9
9+7 = 16

In short let B0 represent the first element in the sequence 0,
let B1 represent the second element, and Bn represent the (n-1)th element in the sequence.

n^2 = Bn = B(n-1) + 2n -1 (This can be proven by induction)

Using this rule for the sequence of perfect squares we can now see that a pattern develops with regard to the remainders after division by 6. (I must credit WritersBlock here. I've never heard of the rule about sqares of primes expressed as 6n+1. I never would have looked at the remainders as a point of interest had I not seen the previous post.)

0/6 = 0 r 0-----1/6 = 0 r 1-----4/6 = 0 r 4-----9/6 = 1 r 3-----16/6 = 2 r 4------25/6 = 4 r 1
36/6 = 6 r 0---49/6 = 8 r 1----64/6 = 10 r 4---81/6 = 13 r 3---100/6 = 16 r 4--121/6 = 20 r 1
...

This pattern is guaranteed to continue because we're adding consecutive odd numbers and every six consecutive odd numbers will add a multiple of six. So, whatever the remainder is for any given number, it will be the same for the sixth number after it.
Proof: Let k represent any odd number.
add k and the 5 consecutive odd numbers after k.
k + k+2 + k+4 + k+6 + k+8 + k+10 =
6k + 30 = 6(k+5)

Using the pattern of remainders {0, 1, 4, 3, 4, 1}, only the second and fifth columns could possibly be the sqares of prime numbers.
If remainder is 0 then A^2 is divisible by 6 which means that A is not prime.
-----(if A is prime then the prime factors of A^2 is A*A. 6 can't be a factor)
If remainder is 4 then A^2 is divisible by 2 which means that A is 2 or "not Prime".
-----(if A is prime then the prime factors of A^2 is A*A. 2 can't be a factor unless A=2)
If remainder is 3 then A^2 is divisible by 3 which means that A is 3 or "not Prime".
-----(if A is prime then the prime factors of A^2 is A*A. 3 can't be a factor unless A=3)
Therefore, if prime numbers exist (and we know they do) then their squares can be expressed as some multiple of six plus 1. Or, A^2 = 6x + 1

A^2 = 6x + 1
A^2 + 26 = 6x + 1 + 26
= 6x + 27
= 3(2x + 9) Which is obviously divisible by 3 and therefore not prime.

• 0

### #86 PGupta

PGupta

Newbie

• Members
• 4 posts

Posted 26 June 2009 - 07:54 PM

aint it goes like a^2 + 5^2 + 1^2 = a*a+26................and sum of square of three prime no. (composite no.) cannot be a prime no.

it can also be said like (a+5+1)^2 - 10a - 2a - 10 which further doesn't turn to be a prime no.
• 0

### #87 jerbil

jerbil

• Members
• 208 posts

Posted 03 August 2009 - 11:00 AM

I think I might have it, straining hard to remember high school algebra and primes. Ok... all primes above ... seven I think ... can be notated as 6n+1 or 6n-1. If you square these, then you have 36n+12n+1 and 36n-12n+1. Right? So you factor these down and you get 12n(3n+1) and 12n(3n-1).

Umm... brain is smoking now, but let's keep at it....

So, take 12n(3n+1). IF n is odd, then 3n+1 must be even, if 3n+1 is odd, then n must be even. So, 12 times an even number is divisible by 24, so 24 is a factor of every prime over 7 (notated as 6n+1) when it is squared. If you add 24 to that, you still have a non-prime. OK?

So, take 12n(3n-1). Ah HA! Same thing. N or 3n-1 must be even, so 24 will be a factor of every prime notated as 6n-1 so we have the same result.

So now we just have to figure out primes less than 7.

7: 49+24 = 73 = PRIME! DAMN! Should have done that first ...

5: 25+24 = 69 = not prime.

3: 9+24 = 33 = not prime.

1: 25 = not prime.

Seven proves it's not true. But, for all primes except for 7 it's true.

Not quite sure why you "think" that 7 (or 5 for that matter) is not a member of the class of 6n +/- 1.
• 0

### #88 ABK

ABK

Newbie

• Members
• 1 posts

Posted 14 September 2012 - 03:02 PM

My first approach was to consider modulo 6.

Assume a > 6, Since a is prime we have a=1, -1 (mod 6)

Thus, a2=1 (mod 6).

We know, 26 = 2 (mod 6)

Therefore a2+26=3 (mod 6)

So any prime greater than 6 would be resulting in a number which is divisible by 3 which NOT prime as required.

So we have to check 2, 3 and 5 which are the only primes less than 6. It is easy to see that each would yield a non-prime in a2+26. So we're done!
• 0

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users