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 post Share on other sites

0 bonanova 84 Posted December 7, 2007 Report Share Posted December 7, 2007 I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime] so Goldbach's "weak" conjecture applies for odd N greater than 5. N = p1 + p2 + p3; [N odd, >5] Let p3=3 [N-3] = p1 + p2; [(N-3) is even, >2] Considering 1 not to be prime, this conjecture fails for 11. 11 = 3 + 3 + 5. Are you objecting to using 3 twice? Quote Link to post Share on other sites

0 Guest Posted December 7, 2007 Report Share Posted December 7, 2007 I just checked and learned that 1 is not considered a prime number [3 is the smallest odd prime] so Goldbach's "weak" conjecture applies for odd N greater than 5. N = p1 + p2 + p3; [N odd, >5] Let p3=3 [N-3] = p1 + p2; [(N-3) is even, >2] Considering 1 not to be prime, this conjecture fails for 11. 11 = 3 + 3 + 5. Are you objecting to using 3 twice? Yes, I thought that was the condition, that p1, p2, and p3 were distinct primes. My apologies if I'm conflating two conversations. Quote Link to post Share on other sites

0 Guest Posted December 7, 2007 Report Share Posted December 7, 2007 Since 'a' is a prime number, without loss of generality we can consider it to be an odd number (the exception is 2, and it is clear that 2*2+26 is not a prime). Then 'a' can be written as "a = 2b+1", for some 'b' a natural number. a*a = (2b+1)*(2b+1) = 4b^2 + 4b + 1. (a*a)+26 = 4b^2 + 4b + 1 + 26 = (4b*(b+1))+27 ...... (*) But to show that (a*a)+26 is not prime, it suffices to show that (4b*(b+1))+27 is a multiple of three. Of course 27 is a multiple of three, so we focus on the second term. But then either 'b' or 'b+1' must be a multiple of three, which is equivalent to say that 'b-1' (OR 'b+2') is not a multiple of three. This is because, for any natural number X, one and only one of 'X', 'X+1', 'X+2' is divisible by three. a = 2b+1 ==> b = (a-1)/2 ==> b - 1 = ((a-1)/2) - 1 = (a-3)/2 .... (**) In order 'b-1' to be a multiple of three, 'a' must be a multiple of three -- which cannot be true because 'a' is prime. Therefore, either 'b' or 'b+1' must be a multiple of three, which implies (4b*(b+1))+27 is a multiple of three. And hence (a*a)+26 is a not prime. If anyone noticed a flaw with the argument, it would be nice to post your comments. I can't immediately spot the problem with your proof, but it must be flawed, because your same reasoning can apply to a*a + 8, which fails at a=3. I forgot to include the condition that 'a' should be greater than 3 so that it can be written in the form of '2b+1' for 'b' a natural number. This is because 'b-1' will be zero if you take a=3. And as we all know zero is a multiple of any number. Even in the original puzzle, if you take a=3, a*a+26 will be 35 which is not a prime number but it is not a multiple of three either. In a sense you can say my argument fails. Thanks for pointing me the error. Quote Link to post Share on other sites

0 bonanova 84 Posted December 8, 2007 Report Share Posted December 8, 2007 Considering 1 not to be prime, this conjecture fails for 11. 11 = 3 + 3 + 5. Are you objecting to using 3 twice? Yes, I thought that was the condition, that p1, p2, and p3 were distinct primes. My apologies if I'm conflating two conversations. In the strong form, the primes are distinct; the weak form [three primes for odd numbers] relaxes that condition. If the 3 can be repeated, then the weak form is established immediately from the strong form. I'm not an expert in number theory. But I found this interesting - that it's universally believed but still not proved. There's more about it here. http://en.wikipedia.org/wiki/Goldbach's_weak_conjecture Quote Link to post Share on other sites

0 Guest Posted December 10, 2007 Report Share Posted December 10, 2007 Yeah, all this kind of smokes me intellectually. I knew there was a reason I went to law school.... Quote Link to post Share on other sites

0 Guest Posted December 10, 2007 Report Share Posted December 10, 2007 I believe it is 7. 7*7= 49. 49+26= 75. correct? 75 isn't prime right? There might be more solutions but i stoped when I got one solution. Quote Link to post Share on other sites

0 Guest Posted December 11, 2007 Report Share Posted December 11, 2007 Since 'a' is a prime number, without loss of generality we can consider it to be an odd number (the exception is 2, and it is clear that 2*2+26 is not a prime). Then 'a' can be written as "a = 2b+1", for some 'b' a natural number. a*a = (2b+1)*(2b+1) = 4b^2 + 4b + 1. (a*a)+26 = 4b^2 + 4b + 1 + 26 = (4b*(b+1))+27 ...... (*) But to show that (a*a)+26 is not prime, it suffices to show that (4b*(b+1))+27 is a multiple of three. Of course 27 is a multiple of three, so we focus on the second term. But then either 'b' or 'b+1' must be a multiple of three, which is equivalent to say that 'b-1' (OR 'b+2') is not a multiple of three. This is because, for any natural number X, one and only one of 'X', 'X+1', 'X+2' is divisible by three. a = 2b+1 ==> b = (a-1)/2 ==> b - 1 = ((a-1)/2) - 1 = (a-3)/2 .... (**) In order 'b-1' to be a multiple of three, 'a' must be a multiple of three -- which cannot be true because 'a' is prime. Therefore, either 'b' or 'b+1' must be a multiple of three, which implies (4b*(b+1))+27 is a multiple of three. And hence (a*a)+26 is a not prime. If anyone noticed a flaw with the argument, it would be nice to post your comments. I can't immediately spot the problem with your proof, but it must be flawed, because your same reasoning can apply to a*a + 8, which fails at a=3. I forgot to include the condition that 'a' should be greater than 3 so that it can be written in the form of '2b+1' for 'b' a natural number. This is because 'b-1' will be zero if you take a=3. And as we all know zero is a multiple of any number. Even in the original puzzle, if you take a=3, a*a+26 will be 35 which is not a prime number but it is not a multiple of three either. In a sense you can say my argument fails. Thanks for pointing me the error. It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met: x > 0 (just to be safe, this may not be necessary), (x+1)%3 = 0 (e.g. 27 from the original) 2*2+x is not prime (since this case was proved by example) 3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1) This appears sound to me right now, but it's starting to get late. Let me know if it checks out. Quote Link to post Share on other sites

0 Guest Posted December 11, 2007 Report Share Posted December 11, 2007 It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met: x > 0 (just to be safe, this may not be necessary), (x+1)%3 = 0 (e.g. 27 from the original) 2*2+x is not prime (since this case was proved by example) 3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1) This appears sound to me right now, but it's starting to get late. Let me know if it checks out. It is correct. Quote Link to post Share on other sites

0 Guest Posted January 27, 2008 Report Share Posted January 27, 2008 Thinking along the same lines as Cawebster, 3 IS the magic number. Here is a proof of the problem: for all primes "a" except 3, a^2+26 is divisible by 3 Part I - Testing when a=2 and when a=3 let a=2 2^2+26= 30 = Composite let a=3 3^2+26= 35 = Composite Part II - All primes a except 3, a^2+26 is divisible by 3 When a number is divided by 3, the remainder can be either 0,1, or 2. As such, any positive integer greater than 2 can be written in the following forms: m=some non-negative integer 3m 3m+1 3m+2 If a is a prime number unequal to three (we already solved the case when a=3 and a=2) then a can be written as 3m+1 or 3m+2. The case when a=3m+1 a^2+26 = (3m+1)^2+26 = 9m^2+6m+27 = 3(3m^2+2m+9) Because this is divisible by three, then when a=3m+1 the result is COMPOSITE The case when a=3m+2 a^2+26 = (3m+2)^2+26 = 9m^2+6m+30 = 3(3m^2+2m+10) Because this is divisible by three, then when a=3m+2 the result is COMPOSITE QED Quote Link to post Share on other sites

0 Guest Posted February 1, 2008 Report Share Posted February 1, 2008 The answer is: 2=a 2*2+26=30 30 factors are: 1,2,3,5,6,10,15,30 A prime number is only divisible by 1 and itself. 2 is the smallest and only even prime number. The only answer can be 2 otherwise it equals a prime number. Quote Link to post Share on other sites

0 Guest Posted February 2, 2008 Report Share Posted February 2, 2008 The answer is: 2=a 2*2+26=30 30 factors are: 1,2,3,5,6,10,15,30 A prime number is only divisible by 1 and itself. 2 is the smallest and only even prime number. The only answer can be 2 otherwise it equals a prime number. If only it were that simple... but as I have demonstrated in my proof: 3*3+26= 35 and 35 is composite therefore the conjecture that the only answer is 2 is incorrect. Quote Link to post Share on other sites

0 Guest Posted March 5, 2008 Report Share Posted March 5, 2008 (edited) ********************************** The theory is incorrect. a*a+26 can be prime if 'a' is prime. Ex when a=11, a*a+26 is 137 which is a prime number. Edited March 5, 2008 by amateur Quote Link to post Share on other sites

0 Guest Posted March 5, 2008 Report Share Posted March 5, 2008 ********************************** The theory is incorrect. a*a+26 can be prime if 'a' is prime. Ex when a=11, a*a+26 is 137 which is a prime number. sorry my mistake 11*11+26 is 147 Quote Link to post Share on other sites

0 Guest Posted March 5, 2008 Report Share Posted March 5, 2008 sorry my mistake 11*11+26 is 147 .... and 147 is not prime. Therefore, the theory is correct, right? Quote Link to post Share on other sites

0 Guest Posted March 5, 2008 Report Share Posted March 5, 2008 Any number (a) times itself creates an even number added to an even number (26) will always be divisible by 2, by meaning NOT PRIME. did I get it right? Quote Link to post Share on other sites

0 Guest Posted March 5, 2008 Report Share Posted March 5, 2008 Sorry, I am WEtodd.... I revoked my last statement i thought it was a+a Quote Link to post Share on other sites

0 Guest Posted March 6, 2008 Report Share Posted March 6, 2008 Any number (a) times itself creates an even number added to an even number (26) will always be divisible by 2, by meaning NOT PRIME. did I get it right? Ny number times itself is even???? like..3*3=9 (even?), 5*5=25(even?),......your math is off i think Quote Link to post Share on other sites

0 Guest Posted March 6, 2008 Report Share Posted March 6, 2008 To do this proof, we need to show that for any prime number 'a', ((a*a) + 26)/3 is an integer. We do this by the modulus of each number. 26 is congruent to 2 mod 3. Which means if 26 is divided by 3, the remainder is 2. It then suffices to show that a*a is congruent to 1 mod 3. Which means if a*a is divided by 3, the remainder is 1. First, we know that 3 does not divide 'a' because 'a' is prime. Then, Suppose the prime 'a' is congruent to 1 mod 3. We can write 'a' as 3n + 1 where 'n' is a natural number (integer > 0). Then (3n + 1)*(3n + 1)=9*(n^2) + 6n + 1 using the FOIL method. It is easy to see that '9*(n^2)' can be divided by 3 as can '6n'. But 1 is still congruent to 1 mod 3. This means we can rewrite the sum as a*a=3k + 1 where 'k' is a natural number. So if 'a' is congruent to 1 mod 3, so is 'a*a'. Then, Suppose the prime 'a' is congruent to 2 mod 3. We can write 'a' as 3n + 2 where 'n' is a natural number (integer > 0). Then (3n + 2)*(3n + 2)=9*(n^2) + 12n + 4 using the FOIL method. It is easy to see that '9*(n^2)' can be divided by 3 as can '12n'. This is the same as before, but with one extra step. We see that 4 is congruent to 1 mod 3. This means we can rewrite the sum as a*a=3k + 1 where 'k' is a natural number, even if 'a' is congruent to 2 mod 3! In conclusion, 'a*a' always has a remainder of 1 when divided by 3, and 26 has a remainder of 2 when divided by 3. This means when the remainders are added together, they are ALWAYS divisible by three, thus it cannot be a prime. This method works for any number like 26, where it is congruent to 2 mod 3. So 'a*a + 2', 'a*a + 5', 'a*a + 212' are all not primes, as long as 'a' is prime. -Mathematics Graduate Student University of Missouri Quote Link to post Share on other sites

0 Guest Posted March 7, 2008 Report Share Posted March 7, 2008 In conclusion, 'a*a' always has a remainder of 1 when divided by 3, and 26 has a remainder of 2 when divided by 3. This means when the remainders are added together, they are ALWAYS divisible by three, thus it cannot be a prime. This method works for any number like 26, where it is congruent to 2 mod 3. So 'a*a + 2', 'a*a + 5', 'a*a + 212' are all not primes, as long as 'a' is prime. Correct, with one condition though. 'a' should be greater than three. The reason can be found in the previous threads. For a=3, 3*3+26=35 which is not divisible by three despite 'a' being a prime number. Quote Link to post Share on other sites

0 Guest Posted March 16, 2008 Report Share Posted March 16, 2008 (edited) If "a" is a prime number, then prove, that a*a+26 is NOT a prime number (whether it is true). Me being who I am, I took the liberty of writing out some of the equations Prime numbers -- Soultion-Prime number factor 2 -- 30 Composite -(2) 3 -- 35 Composite - (5) 5 -- 51 Composite - (3) 7 -- 74 Composite - (2) 11 -- 147 Composite - (3) 13 -- 195 Composite - (3) 17 -- 315 Composite- (5) 19 -- 387 Composite- (5) Unless otherwise, I believe that all prime numbers squared, plus 26 is a composite number. Thus concluding that the answer would be: All prime numbers. Edited March 16, 2008 by Standards Quote Link to post Share on other sites

0 Guest Posted March 16, 2008 Report Share Posted March 16, 2008 Me being who I am, I took the liberty of writing out some of the equations Prime numbers -- Soultion-Prime number factor 2 -- 30 Composite -(2) 3 -- 35 Composite - (5) 5 -- 51 Composite - (3) 7 -- 74 Composite - (2) 11 -- 147 Composite - (3) 13 -- 195 Composite - (3) 17 -- 315 Composite- (5) 19 -- 387 Composite- (5) Unless otherwise, I believe that all prime numbers squared, plus 26 is a composite number. Thus concluding that the answer would be All prime numbers. It is true that if "a" is a prime number, then "a*a+26" is not a prime. But the method of proof you followed is not correct. Look at the following conjecture. For any positive integer 6n-1 or 6n+1 (or both) is a prime. According to your proof method: 1 -- 5 Prime (6n-1) 2 -- 11 Prime (6n-1) 3 -- 17 Prime (6n-1) 4 -- 23 Prime (6n-1) 5 -- 29 Prime (6n-1) 6 -- 37 Prime (6n+1) ... here 6n-1 is 35 which is a composite number. 7 -- 41 Prime (6n-1) 8 -- 47 Prime (6n-1) But the conjecture is not correct. Take n=20. 6*20+1 = 121, which is divisible by 11. 6*20-1 = 119, which is divisible by 7 (and 17). So, I suggest you better use one of the well known mathematical proof methods. Quote Link to post Share on other sites

0 Guest Posted March 16, 2008 Report Share Posted March 16, 2008 But, we're trying to prove how the expression (a*a)+26 can not equal another prime number, not how we find a prime number. Quote Link to post Share on other sites

0 Guest Posted March 16, 2008 Report Share Posted March 16, 2008 But, we're trying to prove how the expression (a*a)+26 can not equal another prime number, not how we find a prime number. I know, and said that in my post. But I am commenting on YOUR PROOF METHOD. I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them. I was just trying to demonstrate such kind of proof method doesn't want all the time. Quote Link to post Share on other sites

0 Guest Posted March 19, 2008 Report Share Posted March 19, 2008 I know, and said that in my post. But I am commenting on YOUR PROOF METHOD. I cannot proof a conjecture by exhaustion because there are infinite numbers, and it is impossible to test all of them. I was just trying to demonstrate such kind of proof method doesn't want all the time. It's prove, not proof, in your third sentence. And work, instead of want, last sentence. But other than that I understand. Quote Link to post Share on other sites

0 Guest Posted March 19, 2008 Report Share Posted March 19, 2008 To do this proof, we need to show that for any prime number 'a', ((a*a) + 26)/3 is an integer. We do this by the modulus of each number. 26 is congruent to 2 mod 3. Which means if 26 is divided by 3, the remainder is 2. It then suffices to show that a*a is congruent to 1 mod 3. Which means if a*a is divided by 3, the remainder is 1. ...... blah blah blah -Mathematics Graduate Student University of Missouri Here is another way to look at this proof: The proof requires bases. I will be working in base three (number set 0, 1, 2) Here is a base 3 multiplication table (leaving out 0) 1*1=1, 1*2=2, 2*2=11. So, any number divisible by 3 will end in 0 (base 3). A prime number is not divisible by 3, so it will end in either 1 or 2. When I square that prime number, it will end in 1, always. (see multiplication table). So, I just need to add 2, or 3n+2 to the square to make it divisible by 3. I can use 3n + 2 in the place of "26" to make this puzzle work. So that the addend is not obvious, n is even. Therefore, the problem works for a*a+2 as well as a*a+38 Quote Link to post Share on other sites

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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).

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