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        riddles >> medium >> system of equations
        
(Message started by: Christine on Apr 10th, 2014, 3:23pm)
Title: system of equations
Post by Christine on Apr 10th, 2014, 3:23pm
I ran a program to find solutions to

A^3 - B^3 = x^5
A^5 - B^5 = y^3

but couldn't find any.
Perhaps somebody would like to take a shot at this.

Could anyone offer an analytical solution?
Title: Re: system of equations
Post by pex on Apr 11th, 2014, 4:35am
I just wanted to share that Wolfram Alpha is spectacularly unhelpful in this case.
Title: Re: system of equations
Post by Christine on Apr 11th, 2014, 8:05am
Yeah, no kidding. I've tried Wolfram, too, at first.
Title: Re: system of equations
Post by SWF on Apr 17th, 2014, 9:15pm
That is a nice problem. I assume you are looking for the A, B, x, and y all positive integers, otherwise there are obvious trivial solutions when one or more of them are zero. There are an infinite number of non-trivial solutions too. The lowest I have found is:

A=562265004868829486
B=281132502434414743
x=43487680369
y=378999812623353224604123685177
Title: Re: system of equations
Post by Christine on Apr 21st, 2014, 2:38pm

on 04/17/14 at 21:15:22, SWF wrote:
I assume you are looking for the A, B, x, and y all positive integers


Yes. Sorry. I forgot to mention that.


Quote:
......There are an infinite number of non-trivial solutions too. The lowest I have found is:

A=562265004868829486
B=281132502434414743
x=43487680369
y=378999812623353224604123685177


Thanks for providing an example.
Title: Re: system of equations
Post by SWF on Apr 22nd, 2014, 7:49pm
Here is the analytical approach I used to come up with those numbers. Start with two integers c and d, such that their difference is the 15th power of some integer:
c-d=z15

The difference of cubes or 5th powers of c and d both can be factored with c-d a factor. Define integers N and M by:
N=(c3-d3)/(c-d)
M=(c5-d5)/(c-d)

If you choose A and B as (p and q are any non-negative integers):
A=c*N15p+3M15q+10
B=d*N15p+3M15q+10

Then
A3-B3=z15N45p+10M45q+30 (a perfect 5th power)
A5-B5=z15N75p+15M75q+51 (a perfect cube)

The solution given previously is for c=2, d=1, p=0, q=0.
Title: Re: system of equations
Post by Christine on Apr 24th, 2014, 7:58pm
Awesome!


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