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        general >> wanted >>  help i cannot figure this prob out 
        
(Message started by: alton pezzoli on Nov 19^th, 2005, 7:10pm)

Title: help i cannot figure this prob out
Post by alton pezzoli on Nov 19^th, 2005, 7:10pm

solve this equation and find all answers in natural numbers: X^3+3=4y(y+1)

[hideb][/hideb]Subtract 3 and factor to get x^3=(2y+3)(2y-1). divided the difference, 4, bec each is odd the gcd =1. So each is a perfect cube. Let
2y+3=a^3, 2y-1=b^3 so that a^3-b^3=4. this is all i understand. This problem is posted on
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61009
pls help me get the rest [hideb][/hideb]

my login is not working

Title: Re: help i cannot figure this prob out
Post by alton on Nov 19^th, 2005, 7:53pm

oh also
I am having problems compiling documents in Latex
i.e. the build output will not let me click on it
Does anyone know what I am doing wrong (it’s probably something stupid)

Title: Re: help i cannot figure this prob out
Post by Eigenray on Nov 19^th, 2005, 9:10pm

No two cubes of integers can differ by 4.

To go into more detail, since a³ = 2y+3 >= 5, we have a >=2. And since b<a, in fact b <= a-1, and therefore
a³-b³ >= a³-(a-1)³ = 3a(a-1)+1 >= 7 > 4.

What I don't understand is what "the build output will not let me click on it" means.

Title: Re: help i cannot figure this prob out
Post by ap on Nov 20^th, 2005, 10:27am

divides the difference
D (2y+3) - (2y-1)
D 4

a^3 - b^3=4

you cant do that substitution?

Title: Re: help i cannot figure this prob out
Post by ap on Nov 20^th, 2005, 11:07am

a^3-b^3 >= a^3-(a-1)^3 = 3a(a-1)+1 >= 7 > 4. That’s exactly what they got on Aops but they also have

[hideb][/hideb] a^3-(a-1)^3 > a^3 - b^3
b^3 > (a-1)^3
b> a-1
But for a^3-b^3 > 0, b<a then b must be between 2 consecutive integers, a contradiction. Therefore no solution [hideb][/hideb]

How did you get >=7>4 where did the 7 come from pls explain

Title: Re: help i cannot figure this prob out
Post by Eigenray on Nov 20^th, 2005, 3:05pm

If a >= 2 and b<a are integers, then
a³-b³ >= 2³-1³ = 7 > 4,
as 1 and 8 are the closest two (positive) cubes.

on 11/20/05 at 10:27:59, ap wrote:

you cant do that substitution?

Eh?

Title: Re: help i cannot figure this prob out
Post by ap on Nov 20^th, 2005, 5:19pm

oh ;D i finally get it thanks
I should have looked at the solution more ::)