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Math Help - [SOLVED] Modified Wallis Example

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    [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!
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    Quote Originally Posted by Samson View Post
    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.
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    Quote Originally Posted by undefined View Post
    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?
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    Quote Originally Posted by Samson View Post
    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.
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    Quote Originally Posted by undefined View Post
    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)
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    Quote Originally Posted by Samson View Post
    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.
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    Thank you undefined!
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