chk out these mathematical curiosities

[Shalmanese]

okay I need the solutions to these 2 problems

1. find 2 cubics which, added together, make another perfect cubic.

2. find an even number apart from 2 which is not the sum of two primes

first ones to get these wins a 128MB stick of micron RAM.

feel free to discuss it on this thread but PM me with the answer otherwise you dont get the prize.

edit: title changed as per pamchenko's request

 

[Javelin]

Obviously both questions have no solution. The first one, i believe is some variant on Fermat's Last Theorem. If it is then the proof will be very comlicated and way beyond anything I can do.

As for the second one, it might be easier to prove. BTW, 2 is not a sum of two primes(0 and 1 are not prime).

 

[Shalmanese]

<< 2. find an even number apart from 2 which is not the sum of two primes >>



I never said 2 was the sum of 2 primes

I will seriously buy someone a 128MB stick of micron RAM if they solve the second one

 

[shifrbv]

1)How about: 677959805103821424723263992665061838773573375138707 37934706199386093375292356829747318657796585767362

Don't forget to send simms

 

[IamDavid]

Cheater shifrbv, I was gonna use that answer...

 

[Javelin]

Here is a non-formal, intuitive proof of (2).

The even number X must be the sum of 2 odd numbers as even numbers are prime and cannot be used.

So if (X-1,1) is a solution then X-1 must be non-prime. Same for X-3, X-5, X-7... X-9 can be prime, X-11 must be non-prime etc.

This means that there needs to be symettry... primes must be matched by non-primes.

Although, I cannot prove it, the number sequence for primes has no pattern or symettry necessary to make this problem work.

 

[thEnEuRoMancER]

The first problem (Fermat's Last Theorem) has been solved a couple of years ago by Andrew Wiles.

The second problem (The Goldbach conjecture)is yet to be solved.

And you, Shalmanese, are a troll, because there's a 1 million $ reward for solving the Goldbach conjecture.

 

[br0wn]

Solution for no 1 :
Read the 150 pages proof by Andrew Wiles.

Ok, where is my 128MB ram ?

 

[Shalmanese]

hmm.. okay, my money making schemes have failed.

here is a genuine one though but no prize

find a perfect square that, when doubled leads to another perfect square or prove that this is impossible.

 

[Wah]

<<And you, Shalmanese, are a troll, because there's a 1 million $ reward for solving the Goldbach conjecture. >>

lol..

 

[pamchenko]

shalmanese you think you are really smart by posing these questions? I bet you don't know real mathematics and pose like you do by offering these questions...why don't u just stop wasting everyone's time and post &quot;chk out these mathematical curiosities&quot; instead of neffing &quot;can someone help me solve these two simple...&quot; simple... f*** you...who do you think you are

 

[pamchenko]

I'm just so mad at you because I came into this thread looking to help out some HS kid leaving his HW to the last minute (something i am guilty of)...but instead I find some guy being stupid and deceitful

 

[Fierysonic]

relax, dude

 

[Shalmanese]

I am sorry pamchenko for causing you so much grief, I seriously did not think that anybody would be offened by this kind of thing. It was late and I was bored. I know thats no excuse but I never thought someone could get so p*ssed off about something like this

Okay, the thread title has been changed

 

[Shalmanese]

And heres another genuine problem for you all just to make it up

(x-a)(x-b) can be expanded into x^2 - ax - bx + ab
(x-a)(x-b)(x-c) can be expanded into (x^2 - ax - bx + ab)(x-c)
which can be fully expanded into x^3 - ax^2 - bx^2 + abx - cx^2 +acx +bcx -abc

what is (x-a)(x-b)... (x-z) fully expanded

 

[Shalmanese]

bump

 

[mAdD INDIAN]

For the last question..about (x-a)(x-b)...(x-z)...isn't there a Fermats triangle for that?

basically it's
1
1 1
1 2 1
1 2 2 1
1 2 3 2 1

something like that..

so it could be like this x^4 + 2yx^3 + 3 y^2x^2 + 2xy^3 + y^4

something like that...not exactly for the one u said..

 

[Zedfu]

mAdD INDIAN, i think you mean pascal's triangle.
and you mean it's
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
and so on....

 

[strawberry]

Reply to Shalmanese: The answer is 0. At the end of the sequence, you get (x-x), which is 0, so the whole thing is 0. Very cute.

 

[Mday]

i had a hard enough time figuring out the primes of gaussian integers =(

 

[thEnEuRoMancER]

Pascal's triangle is used to determine the terms a(k,n) in expansion of a binome:

(x+y)^n = sum(k=0...n) ( a(k,n)*x^(n-k)*y^k ) ,

a(k,n) = n!/(k!*(n-k)!)

 

[Shalmanese]

arghh!! stupid me!

If you dont know what I am talking about see below

why cant they let you delete your own posts?

 

[Shalmanese]

umm.. reverse triple post, see below

 

[Shalmanese]

Strawberry: good job, generally the more &quot;mathematical&quot; a person is, the quicker they go for the binomial expnasion route which gets them nowhere .

Nobody has solved the first one yet but I will give you a clue:

A person once (supposedly) died when trying to prove a derivative of this problem

Mday: all of the gausiann primes 0>x>10 and 0>y>10 here for you:
1+i, 2+i ,3 ,3+2i ,4+i ,5+2i ,7+i ,5+4i ,7 ,7+2i ,6+5i ,8+3i ,8+5i ,9+4i

to find a Gaussian prime, x + iy, (x + iy)*(x-iy) has to equal a prime

Hope that is a help to you

 

[Shalmanese]

reverse triple post