Guest Posted November 17, 2007 Report Share Posted November 17, 2007 If "a" is a prime number, then prove, that a*a+26 is NOT a prime number (whether it is true). Quote Link to comment Share on other sites More sharing options...

0 Guest Posted December 10, 2008 Report Share Posted December 10, 2008 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 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted December 19, 2008 Report Share Posted December 19, 2008 a=2 (2*2)+ 26=30 4+26=30 30=30 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 13, 2009 Report Share Posted February 13, 2009 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 Quote Link to comment Share on other sites More sharing options...

0 Prime Posted February 13, 2009 Report Share Posted February 13, 2009 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. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 16, 2009 Report Share Posted February 16, 2009 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 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 26, 2009 Report Share Posted February 26, 2009 just to let everyone know i didnt read pages 2-8 so if this was put up then sry 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 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 26, 2009 Report Share Posted February 26, 2009 (edited) Wow, so may formulas... Edited February 26, 2009 by paperphoenix Quote Link to comment Share on other sites More sharing options...

0 Guest Posted March 23, 2009 Report Share Posted March 23, 2009 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? Quote Link to comment Share on other sites More sharing options...

0 Guest Posted June 26, 2009 Report Share Posted June 26, 2009 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. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted June 26, 2009 Report Share Posted June 26, 2009 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. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 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. Quote Link to comment Share on other sites More sharing options...

0 ABK Posted September 14, 2012 Report Share Posted September 14, 2012 My first approach was to consider modulo 6. Assume a > 6, Since a is prime we have a=1, -1 (mod 6) Thus, a^{2}=1 (mod 6). We know, 26 = 2 (mod 6) Therefore a^{2}+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 a^{2}+26. So we're done! Quote Link to comment Share on other sites More sharing options...

0 Bella Lane Posted April 9, 2023 Report Share Posted April 9, 2023 "a" =3 3*3=9, and 9+26=35 35 is NOT a prime number. In conclusion, 3 is a possible answer. Quote Link to comment Share on other sites More sharing options...

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