#proof

81 messages · Page 1 of 1 (latest)

hollow zealotBOT
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urban blaze
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not sure how to start

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i know that x, y, and z have to hav edifferent signs

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they cant all be positive or all be negative

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has to be some mix of + and -

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i tried combining the fractions and cross multiplying but i dont see ideas

keen flame
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Hm... Well, we can try multiplying everything by xyz(x + y + z):
(x + y + z)(xy + yz + xz) = xyz
x^2 y + xy^2 + y^2 z + yz^2 + x^2 z + xz^2 + 2xyz = 0
Maybe we can try treating, say, z as a parameter, and bring the resulting quadratic polynomial in x, y to the canonical form?

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Ah, wait, no.

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It isn't quadratic in x and y.

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Then I don't know.

fierce rock
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It's quadratic in z. We can solve for z in terms of x and y with the quadratic formula.

keen flame
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Oh. In that case, we can just, say, solve for z. It is quadratic in just z, after all.

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Yeah, right!

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Oh, wait a minute. Doesn't it kinda look like ||(x + y)(y + z)(z + x) = 0||?

fierce rock
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And at that point we can simplify things by checking if the discriminant is a perfect square.

fierce rock
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So you just want to prove whether it's possible for the equation to be true, right?

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In which case you don't need all solutions, just a solution.

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Or a proof of no solutions.

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But a solution is sufficient to prove that a solution exists.

urban blaze
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Yes

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Oh

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Is it z^2 (x+y) + z(x^2 + y^2) + (x^2 y + y^2 x + 2xyz)

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for a, b, and c

fierce rock
urban blaze
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but x^2 + 2xy + y^2 is (x+y)^2

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so z^2 (x+y) + z(x+y)^2 + (x^2 y + y^2 x)

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x^2 y + y^2 x is xy(x+y)

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im getting z = -y, or z = -x

keen flame
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Yeah, that's correct.

urban blaze
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o

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1/x + 1/y + 1/z = 1/(x+y+z)

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if z is -y

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its just 1/x = 1/x

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if z is -x

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its 1/y = 1/y

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x+z=0

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y+z =0

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if i want to find the number of solutions

keen flame
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Remember that you can't have any of the variables or their sum equal to 0.

urban blaze
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oh right

urban blaze
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just not 3

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or else 1/(x+y+z) is undefined

keen flame
urban blaze
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for x+z=0
there should be solutions

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20 *

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and for each of those solutions

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y is from -10 to 10 but not 0

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so for each of those solutions there are 20 possible y's

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so 400

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wait

keen flame
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I don't think there are that many. Some may be double-counted.

urban blaze
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oh

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this is the whole question btw

be58aa6bdc7c71ee103d664eb1882094.png
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just had to prove this earlier part to work toward a solution for the problem

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if z = -y

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1/x = 1/w

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then for our solution

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if we make w be x since they will be the same number anyway

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x,y,-y,x

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think we can just ignore the x at the end since it will not matter since its the same as the first x

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x,y,-y

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-y depends on y

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it will always be the - of y

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x,y

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if x has 20 and y has 20 then there are 400.. like

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wait

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yea..

fierce rock
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Because you don't necessarily only get solutions when two of the terms cancel.

urban blaze
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but from our quadratic didnt we prove we can only get solutions when z = -x,-y (so two terms cancel)?

fierce rock
urban blaze
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oh

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sorry cryboss

fierce rock
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Next time, ask the question you're actually trying to answer.

urban blaze
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okay

urban blaze
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solved

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+close