Need maths help

Is it possible to make x^2 - y^3 a product of 2 or 3 expressions? Meaning:

(     )*(     ) = x^2 - y^3

(     )*(     )*(     ) = x^2 - y^3

I can't find that...

Easily. But not at all if you don't allow fractional powers. The proof of this needs something like the theorem "The ring of polynomials in two variables with rational powers is a unique factorisation domain". This is probably undergraduate algebra.

Since x^2-y^3 = (x - y^(3/2)) * (x+y^(3/2))

If you wanted to write it as a product of two or more factors, you'd have to factor either of these factors. Since each of these factors is of the form x - Q(y), this is not going to help.

Realising that the theorem I stated is not so obvious is a point that could be missed. In fact in a different context, a great historical mathematician made this assumption incorrectly when working with a different ring.

Quiz - to which mathematician am I referring?

We could treat this as a difference of two squares.  We do not typically treat the y^3 term this way, since it is not technically a "perfect square."  Recall that (a+b)(a-b) = a^2 - b^2.  In your case a = x, since a is equal to the square root of a^2, and a^2 is x^2, and the square root of x^2 is x.  The question must be then what is your term b?  Your b-term would be the square root of y^3.  Therefore with a difference of squares we have x^2-y^3 = (x + root[y^3])(x - root[y^3]) in general.  Using a bit more cosmetic notation, we can express this expression as (x + y^3/2)(x - y^3/2).  As far as expression x^2-y^3 as a product of three terms, I'm not quite sure off of the top of my head.  Hope that helps.

MM

yes but only with natural powers...

Elroch, go easy on Kacparov! Kacparov, MM's (and Elroch's, of course) solution looks fine to me because rational numbers (y to the 3/2) are a subset of the natural numbers-so it's a natural power. I don't think you have to worry.

wel it was part of a bigger problem (the one I had problems with) and it doesn't work...

Well, the definitive answer is simply that there is no factorisation with natural number powers. Strictly speaking one does have to check the lemma I mentioned to prove this, even though it is natural to assume it's true. Apologies to Kacparov for throwing in some maths that requires a course or two of university level algebra.

The answer to my earlier question was the incomparable Leonard Euler. The story of Euler's mistake is a very interesting one. For one of the prolific mathematicians of all time, the fact that this is referred to in the singular is very enlightening!

[Incidentally, I suppose strangequark meant to say rationals are a superset of the natural numbers, not a subset.]

Ok, guess I have to solve the problem in some other way...

Yes, as Sherlock Holmes said "when you have eliminated the impossible, whatever remains, however improbable, must be the truth"