# Thread: [SOLVED] Modified Wallis Example

1. ## [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!

2. 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 $\displaystyle x \le y$.

3. 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 $\displaystyle x \le y$.

Is it required that $\displaystyle x \le y$ ? That is the only solution possible? How did you reach that conclusion?

4. 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 $\displaystyle x \le y$ 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.

5. Originally Posted by undefined
I let $\displaystyle x \le y$ 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 $\displaystyle x \le y$ ? (This is just for formality purposes)

6. 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 $\displaystyle x \le y$ ? (This is just for formality purposes)
This is up to preference. Letting $\displaystyle x \le y$ is a mathematical shorthand of sorts, see: Without loss of generality.

7. Thank you undefined!