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[Apolo Program]

Spoiler for Solved!!

1) a * a + 26
= a ^ 2 + 26, easier to read

Being "a" prime:

2) a = 6n + 1

or

3) a = 6n - 1

So take 2) and subtitute on 1) we'll get

(6n + 1) ^ 2 + 26
= (6n) ^ 2 + 2 (6n) + 1 + 26
= (6n) ^ 2 + 2 (6n) + 27
= (6n) [(6n) + 2] + 27
= 3 (2n) { [(6n) + 2] + 9 }

So this number can ve represented by a product of 2 numbers other than itself "a" and "1". Therefore is not prime.

Subtituting 3) we get almos the same:

3 (2n) { [(6n) - 2] + 9 }

Therefore there is no way a * a + 26 is prime

[Udontn0me]

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

Spoiler for mathematical ingenius

a prime number multiplied by itself is never a prime number (expect 1 being its unique) and if you add anything over 1 then you can never get the answer to be a prime number.

i have no idea why everyone else is putting up these long formulas to prove it can't be done

thx for looking at my post

[paperphoenix]

Wow, so may formulas...

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

[Ryan91]

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?

[PGupta]

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.

[PGupta]

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.

[jerbil]

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.

[ABK]

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!