Somewhat disappointed no one took the challenge. This is a beautiful problem that by middle school algebra exposes what real mathematics can be about, proving things for an infinite case. Ok, I will start the problem. First, translate the problem into algebra: X^3 - 1 = p The obvious first thing to do with any number theory problem is to look at factorizations. Set the left hand side equal to zero [this is where technique matters and if you don't know it, on more complicated problems you could not even get started. You gotta know some theory, and that is what an education gets you. The creative part is to push beyond what is known and create theory for others to follow] X^3 - 1 = 0 A solution is 1^3 - 1 = 0 therefore (X - 1) is a factor of (x^3) [extra credit, by what theorem?] Ok that is a clue. Can you solve it from here?
No offense, but the answer to your question is answered by your question itself! So your question is answered by analysing the meta-question. Soon we would fall in infinite regress and get nowhere. I can tell you why I give a shit, but I can't tell you why you should give a shit. I can't even do that for really important things. Just look at all the debate on ET over things that are self-evident with massive amounts of scientific evidence!
Well, that is a question that would get me going for hours and hours. I will give you the really short answer: the way mathematics is taught in school ruins what it is for 99% of us. I am lucky that I can appreciate this proof as much as I can appreciate Beethoven's 9th Symphony, or DaVinci's drawings, or any other Art Form. In fact, slightly more complicated versions of this problem gets one to state of the art research very fast, involving extremely beautiful mathematics. I claim you probably would find the problem beautiful as I do, if only your education were not ruined by the mathematical education you were given. I can't say that for certain, but I am certain that the way mathematics is taught ruins it for most of us. I cannot do better than what Lockhart says.Read the first few paragraphs of, Part I Lamentation, here: http://www.amazon.com/Mathematician...93353207&sr=8-1&keywords=lockhart+mathematics
I suppose one could start by saying: if x is an odd number, then the cube will be an odd number. Every odd number that subtracts one results in an even number. Even numbers are not prime. So, this removes all the odd numbers. Next I have to figure out how to remove all the even numbers. Is this type of story proof acceptable, or does it have to be formulas?
Next, I can remove all the multiples of 10 (10, 100, 100000, etc): (10^y)^3 - 1, for any y, will be a number as 1 with many zeroes behind it. That number subtracting 1 will result in a number that is a bunch of 9's. The sum of the digits of that number will thus be a multiple of nine. All multiples of 9 are divisible by 3. Then, there is another theory that says any number where the digits add up to a number divisible by three, means that number is divisible by 3. Therefore, any multiple of 10 cubed minus 1 will not be a prime number.
You have proven the same thing you proved above. The approach you are taking is probably too hard, but instructional nonetheless. I can dispense with all these arguments of the type you are using by just saying: all primes are congruent to 1 or 3 mod 4. Theorem: All primes are greater than 2 are # 1 or 3 (4) Congruence is an amazing tool invented by Gauss, considered the Prince of Mathematics, but today we could say he was the Mozart of Mathematics. If I say, 5 is congruent to 1 mod 4, it means that if I divide 5 by 4, I get a remainder of 1 because 5 = 4 * 1 + 1 . No fractions, just integer division and remainders. So the above theorem says that all primes leave a remainder of 1 or 3 when divided by 4. This is easily proved, but you can take it as a theorem. This theorem is more general than both of the ones you stated above. The # sign below means "congruent to", and putting a number in parentheses is short for saying "mod whatever is in the paren": "x odd" is the same thing as saying that x # 1 or 3 (4). For example 5 # 1 (4), 7 # 3 (4), 9 # 1 (4) etc on to infinity for all odd numbers. Take each case: a) If x # 1 (4), then [x^3 - 1] # 0 (4) so x can't be the odds that are 1 (4). b) if x # 3 (4), then x is the same thing as -1 (4) ,hence [x^3 - 1] # 2 (4) so x can't be the odds that are # 3 (4) since x comes out to be 0 or 2 (4) in either case of x^3 - 1 with x being odd, and all primes are of the form p = 1 or 3 (4), so x can't be Odd. This is a proof by contradiction. Stated this way, assume x is odd => a contradiction.