xandyare natural numbers.

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- Sep 13th 2008, 04:05 AMcourteous[SOLVED] Difference of cubes

**x**and**y**are natural numbers. - Sep 13th 2008, 04:10 AMMoo
Hello,

That's the sum of cubes (Rofl)

Well, well... The formula is

If you cant remember it, start from and develop it ;)

Solving for it... ? Not impossible.

Note that

Since x and y are natural numbers, and are integers.

So the two factors will be a combination of the divisors of 208. For example 26 and 8 or 13 and 16 or... :)

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Try all the possible values for x+y among 1,2,4,8,13,16,26,52,104,208 (:eek:)

Then note that and is respectively among 208,104,...,2,1

Note that otherwise the product can't be large enough (check it ;))

If you need more explanations, do tell me. But as far as I can see, it will be a long way to do >.> - Sep 13th 2008, 06:10 AMcourteousCorrected - it is a minus...
I've correctly titled the thread(Tongueout), but have mistakenly put

**+ (plus)**instead of**- (minus)**(Doh).

The correct(ed) version:

The solution is . But how do you get it?

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*Moo*, about developing from : what do you do with and ?(Thinking) - Sep 13th 2008, 06:26 AMMooQuote:

*Moo*, about developing from : what do you do with and ?(Thinking)

Develop (x-y)^3 to get

So we know that x and y are positive -----> and

Obviously,

So you have 5 possibilities to test out (x-y stopping at 13).

For example. If then

Are there x and y such that and ? That's a linear system of equations :)

Exactly the same reasoning for the other possibilities ! (Wink)