# [SOLVED] Modified Wallis Example

• Jun 7th 2010, 04:23 AM
Samson
[SOLVED] Modified Wallis Example
Hello all, I have a quick question concering a Wallis Example (click here for background information on Wallis problems). My book states that positive integer solutions exist within many Wallis type examples, followed by a list of them. It does not however list the solutions to them.

I've google searched a good bit and I was able to find how a few problems matched up, but I can't seem to find the solution to (x^3)+(y^3)=35 using methodology that others used to solve their problems.

Can anyone explain how we find all the positive integers solutions to that problem ( x^3 + y^3 = 35 ) ?

All help is appreciated!
• Jun 7th 2010, 04:57 AM
undefined
Quote:

Originally Posted by Samson
Hello all, I have a quick question concering a Wallis Example (click here for background information on Wallis problems). My book states that positive integer solutions exist within many Wallis type examples, followed by a list of them. It does not however list the solutions to them.

I've google searched a good bit and I was able to find how a few problems matched up, but I can't seem to find the solution to (x^3)+(y^3)=35 using methodology that others used to solve their problems.

Can anyone explain how we find all the positive integers solutions to that problem ( x^3 + y^3 = 35 ) ?

All help is appreciated!

I don't understand how number theory really plays into this particular problem. There are very few possibilities. It can be seen with very little work that (x,y) = (2,3) is the only solution for which $x \le y$.
• Jun 7th 2010, 05:00 AM
Samson
Quote:

Originally Posted by undefined
I don't understand how number theory really plays into this particular problem. There are very few possibilities. It can be seen with very little work that (x,y) = (2,3) is the only solution for which $x \le y$.

Is it required that $x \le y$ ? That is the only solution possible? How did you reach that conclusion?
• Jun 7th 2010, 05:08 AM
undefined
Quote:

Originally Posted by Samson
Is it required that math]x \le y[/tex] ? That is the only solution possible? How did you reach that conclusion?

I let $x \le y$ to avoid having to count (x,y) = (2,3) and (x,y) = (3,2) as two separate solutions.

Just plug in numbers. There are very few choices of (x,y) that do not exceed 35.

For example, if (x,y) = (3,4) then x^3 + y^3 is already quite a bit greater than 35.
• Jun 7th 2010, 05:10 AM
Samson
Quote:

Originally Posted by undefined
I let $x \le y$ to avoid having to count (x,y) = (2,3) and (x,y) = (3,2) as two separate solutions.

Just plug in numbers. There are very few choices of (x,y) that do not exceed 35.

For example, if (x,y) = (3,4) then x^3 + y^3 is already quite a bit greater than 35.

Okay, thank you! So if I was to write a formal proof of this, I would write (x,y,)=(2,3) and (x,y,)=(3,2) or would it be more proper to write (x,y)=(2,3) where $x \le y$ ? (This is just for formality purposes)
• Jun 7th 2010, 05:16 AM
undefined
Quote:

Originally Posted by Samson
Okay, thank you! So if I was to write a formal proof of this, I would write (x,y,)=(2,3) and (x,y,)=(3,2) or would it be more proper to write (x,y)=(2,3) where $x \le y$ ? (This is just for formality purposes)

This is up to preference. Letting $x \le y$ is a mathematical shorthand of sorts, see: Without loss of generality.
• Jun 7th 2010, 05:25 AM
Samson
Thank you undefined!