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Hmmm... Is it this easy? The square of any two primes over 7 (if I remember correctly) can be notated as either (6n+1)^2 or (6n-1)^2

This gives 36n^2+12n+1 and 36n^2-12n+1. Both have the +1 as part of it, so if you add 26 then both will have +27.

So using the first you have 36n^2+12n+27 and in the second you have 36n^2-12n+27. Both of these factor by 3, so therefore no prime over 7 when squared and added to 26 can equal a prime.

I just realized that the 6n+1 notation is good for 7 also. As is the 6n-1 for 5.

So 3,2,&1:

(3^2)+26 = 35 Non prime

(2^2)+26 = 30 Non prime

27 = Non Prime

Therefore where A = Prime, A^2+26=Prime is false.

hmmm....excuse me, i just want to point out one thing that i think most of you are wrong about is: that "1" is not a prime number!!!!!

my way for doing this question will be the followings:

A^2+26

then we can have A^2-4+30

(A-2)(A+2)+30

(A-2)(A+2) has to be a non prime number except when A=2,3

and it came out as:

when a=2, then A^2+26=30

a=3, then A^2+26=35

if you test for A^2+26 the answer will be a non-prime number!

therefore: A^2+26 is a non Prime

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If "a" is a prime number, then prove, that a*a+26 is NOT a prime number (whether it is true).

i didn't read all of the posts but i think i have the proof.

if a is prime

prove a^2+26 is not prime

a^2 + 26 = a^2 - 4 +30 = (a+2)(a-2) + 30

a is prime there for it can not be divisible by 3

thus a mod3 = 1 or 2

if a mod 3 = 1 then a +2 is divisible by 3

if a mod 3 = 2 then a - 2 is divisible by 3

thus (a+2)(a-2) is divisible by 3 as is 30

therefore a^2 +26 is divisibly by 3

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i didn't read all of the posts but i think i have the proof.

if a is prime

prove a^2+26 is not prime

a^2 + 26 = a^2 - 4 +30 = (a+2)(a-2) + 30

a is prime there for it can not be divisible by 3

thus a mod3 = 1 or 2

if a mod 3 = 1 then a +2 is divisible by 3

if a mod 3 = 2 then a - 2 is divisible by 3

thus (a+2)(a-2) is divisible by 3 as is 30

therefore a^2 +26 is divisibly by 3

Good proof, but it lacks the final touch. (Same as several other proofs here.)

a=3 Then a is a prime and it is divisible by 3. This case must be tested separately.

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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

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

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.

:P

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

thx for looking at my post :D

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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?

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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.

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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.

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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.

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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!

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