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AlexP



Joined: 14 Nov 2007
Posts: 561
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AlexP
Thought this would be the right place to turn for help with some problems, I've been doing extremelyyyyy simple things like this but this one is drawing a blank for me. Can anyone give me a hint??
Code:
For all integers n > 2 , n^3 - 8 is composite
    

Wow... At first I thought that any number n^3 is even, which would solve it practically by itself, but in a second I realized they weren't. I know it's really simple, but please give me some sort of clue! (Oh yeah, this is not school-related work, so don't insult me for cheating or something.) Probably something simple about factorization, my mind is not getting this one!
Post 05 Feb 2008, 02:17
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LocoDelAssembly
Your code has a bug


Joined: 06 May 2005
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LocoDelAssembly
http://en.wikipedia.org/wiki/Mathematical_induction

The problem of course is that I forgot how to do it Razz

Perhaps I have something on paper, I'll post it if I find it. Since there is few chances to find that info, the other people just don't wait for me and post solution if you know.
Post 05 Feb 2008, 02:44
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AlexP



Joined: 14 Nov 2007
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AlexP
thanks, but i'm trying to get this one proofed, this one just isn't "clicking" in my head... Well, any number >= 2 is a product of powers of primes, I don't think that gets me anywhere. n of course has to be an odd number, I already got that part of it proofed, and I know that when n gets over the first two possible integers the range between prime numbers > 8, but I can't figure out how to prove that. So, is there any prime number that +8 is a perfect cube? I don't think it's possible, maybe... Any help on how to prove that n^3-8 will always be composite?
Hold up, any number that is taken to its square root is always an integer, or it is irrational, so is there a way of spreading it out? IDK, i need help...
Post 05 Feb 2008, 02:52
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bitRAKE



Joined: 21 Jul 2003
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bitRAKE
(-2 + n) (4 + 2 n + n^2) = n^3 - 8

a = n-2
b = n^2 + 2n + 4

a b = n^3 - 8

(a) and (b) are positive integers if (n)>2.
Post 05 Feb 2008, 03:11
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AlexP



Joined: 14 Nov 2007
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AlexP
hmm... or just input 3 into the function n^3 - 8. I just realized that the first possible integer in the equation is the one that solves it Smile. ( actually, contradicts it, but you get it.). 3^3 = 27, - 8 equals 19 which is a prime number.

EDIT: Thanks a lot bitRAKE, that does solve it, but all I needed to do is to prove the resulting integer's compositeness. It would take a little more to prove that both of those variables are not prime, lol I do guess that one of them almost has to not produce a prime, and by multiplying it by the second one you prove the result is composite. Thanks, but apparently there was a number that satisfied the equation and proved an exemption to yours.
Post 05 Feb 2008, 03:18
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LocoDelAssembly
Your code has a bug


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LocoDelAssembly
hahaha, but seems that with n > 3 will work and the reason is because the (n - 2) will be a number > 1 so the result of the multiplication of course is composite.
Post 05 Feb 2008, 03:27
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AlexP



Joined: 14 Nov 2007
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AlexP
uhh you mean (n-2) will be a number > 1, but if you input 4 then the result of n^3 - 8 would be zero. You need n > 4 for that to work, right? I'll submit more later after real homework, fun stuff!
Post 05 Feb 2008, 03:30
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LocoDelAssembly
Your code has a bug


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LocoDelAssembly
x^3 - 8 gives me 56 Confused

And I tested with the calculator (-2 + 4)* (4 + 2*4 + 4*4) to be sure.
Post 05 Feb 2008, 03:35
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AlexP



Joined: 14 Nov 2007
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AlexP
Okay, well that does satisfy the original equation, that all of N > 2 are composite for it, but inputting 3 will fault the theory.... Well, on to the next one... And yes, 4^3 = 64, -8 equals 56, which is composite. I would love to draw up a proof that says 3^3-8 is not composite, I'll quick do that, I don't like to prove ( or disprove) stuff with direct examples of numbers, but here I guess it'll take much longer to find a general proof to counteract it...
Code:
 n  Z, (6(n^2 + n + 1)  (5n^2  3) is a perfect square.
lol that didn't work, here it is:
For all n -in the set- integers, (6(n^2 + N + 1) - (5N^2 - 3) is a perfect square
    

Okay, try and prove this, didn't look at it, just copied from book.
Post 05 Feb 2008, 03:38
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LocoDelAssembly
Your code has a bug


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LocoDelAssembly
Perhaps I'm missing something but isn't proof enough that no matters what is the value of (4 + 2*n + n^2) the result will be always a composite number provided that (n - 2) > 1? If a = (n - 2); b = (n^2 + 2 n + 4) and x = a*b, how could X be prime if it is the result of a multiplication with both parts >= 2? A prime number can't be the result of any a and b pairs.

But again, perhaps I'm missing something about primality of numbers. (And maybe don't understanding your point Razz)

And with the induction principle, at least in the way I used to do it at university, you don't have to disprove by putting an expression, if you find a case where the premise is false you just show it and that is the end of the story.
Post 05 Feb 2008, 03:58
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AlexP



Joined: 14 Nov 2007
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AlexP
lol not the end of the story for me!! Smile Yeah, I got the (n-2) > 1 thing, wrong thing I was looking at. So it will work, good job on the factoring bitRAKE, and please don't worry if these things seem elementary (lol, it's from an elementary number theory book) to you guys because I'm just starting learning NT. Order a few books on ENT, tons of online tutorials on proofing (even a book I found), and I should be off to a good start. never took anything more than Algebra II, (sophomore in high school), so I've never gone into calc or any of that. AP calculus shall be fun in a few years though last year of high school, anyone here take that as a course?

PS: Yeah, didn't even thing of factoring the equation to solve it, but with the primality part of it good job at realizing that Smile

Okay, here's one:

For all integers n, (-1)^n = -1 if and only if n is odd

This is common sense that a negative to an odd power is negative, but how do you prove it? I got to (-1)^2r+1, where 2r+1 is an odd number, but I can't figure out how you would do this. Maybe prove that squaring negative integers results in a positive, but how to translate to general variable for the exponent? Suggestions?
Post 05 Feb 2008, 04:06
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bitRAKE



Joined: 21 Jul 2003
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bitRAKE
(-1)^(2r+1)=(-1)^2r * (-1)
(-1)^2r = (-1)(-1)*(-1)(-1)*... =1

We can remove pairs of (-1) until the exponent is one, this is by definition an odd number exponent: (-1)^5 = (-1)(-1)*(-1)^3 = (-1)(-1)(-1) = (-1)
Post 05 Feb 2008, 04:40
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AlexP



Joined: 14 Nov 2007
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AlexP
yes, so a -mathematical symbol for multiplication "summation"- (II of -1). Good point, I don't know how to proof it though... Maybe just show the first two lines of your equation (I got to that beforeeee..), and then show that 2r as an exponent always produces the same result through some proof I could make, and that all that matters is the +1. Yeah, but how to make that into a robust proof? n ^ 2r = n is what needs to be solved here, nothing more. ( by solved, I mean put into numbers not sentences. And if anyone spells out sentences with hex, I swear I will track you down and thwack you over the head with this computer Smile. So yeah, just solve n^2r = n. I'll do that in school tomorrow if I have time.

and if anyone wants something else, how about this:

There is a unique prime p in the form: n^2 + 2n - 3.

Should only take a few minutes by hand, few milliseconds with code Smile
Post 05 Feb 2008, 04:46
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bitRAKE



Joined: 21 Jul 2003
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bitRAKE
AlexP wrote:
There is a unique prime p in the form: n^2 + 2n - 3
Answer is my favorite number and the day I was born - solved it in my head.
Quote:
a. Any positive integer has a base two form: 1...
b. If an integer is odd then the base two form is: 1...1
c. If (h)=2^a+2^b+2^c+...+1, where a,b,c,...,0 are the exponents for one bits of the binary form of odd integer (h); then (-1)^h = (-1)^(2^a)*(-1)^(2^b)*(-1)^(2^c)*...*(-1)
d. (-1)^2=1
e. Therefor, (-1)^h = -1
...just to relate it to x86 coding, lol.
Post 05 Feb 2008, 06:27
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Xorpd!



Joined: 21 Dec 2006
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Xorpd!
If you want to learn anything about math, forget the high school curriculum which is intended to deaden your consciousness. A couple of lightweight math books are Recreations in Number Theory; the Queen of Mathematics Entertains, Beiler, Dover 1964 and Introduction to Number Theory, Shockley, International Thomson Publishing 1967. For a total of about $20, these may provide the motivation to learn something about math.
Post 05 Feb 2008, 07:41
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AlexP



Joined: 14 Nov 2007
Posts: 561
Location: Out the window. Yes, that one.
AlexP
This isn't part of high-school, our school sucks when it comes to math Smile

Okay, I've got a reallly good one for you guys -
Code:
There exits M > 0, where when n > M the following polynomial produces prime:

n^2 - n + 11

Solution:
http://en.wikipedia.org/wiki/Formula_for_primes

Just research Euler's quadratic and the proof lies there.

    


Here's another, from before. Pretty simple, but may be interesting to some people.
Code:
There exist prime P in the form n^2 + 2n - 3

Solution:
Just factor for (n+3)(n-1), then take each = 0 and solve...

    


Well, no one has been replying, but if you would like more please post! (feel free to post your solutions or thoughts about those two problems, especially the Euler one).
Post 05 Feb 2008, 22:26
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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revolution
This whole topic should be in Heap. There is no programming to see here folks.
Post 06 Feb 2008, 04:10
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LocoDelAssembly
Your code has a bug


Joined: 06 May 2005
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LocoDelAssembly
Moved. Sorry, I didn't realize that it wasn't there already.
Post 06 Feb 2008, 04:35
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MCD



Joined: 21 Aug 2004
Posts: 604
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MCD
Does anyone here have a proof for the following problem in tex?
Code:
\forall n > 2 \land n \in 2\N : \exist p,q \in \P : p+q = n

with \P being the set of prime numbers
    
Post 13 Feb 2008, 21:16
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
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MHajduk
MCD wrote:
Does anyone here have a proof for the following problem in tex?
Code:
\forall n > 2 \land n \in 2\N : \exist p,q \in \P : p+q = n

with \P being the set of prime numbers
    
Hey, MCD, you need a proof of Goldbach's conjecture? As far as I know, it's believed to be true but nobody proved it yet. Wink
Post 13 Feb 2008, 21:31
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