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Solve for system of equations

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Given:
    x7 + y7 =1
    x + y = 1

    Find the integer value(s) of (x-y)2.

    2. Relevant equations
    3. The attempt at a solution
    I thought of substituting for y and then finding the rational roots but then I realized x and y don't have to be rational numbers for (x-y)^2 to be an integer. I am 90% sure the only solution is 1 (when x=1,y=0 or x=0,y=1) but don't know how to prove it. Any hint? Thanks.
     
  2. jcsd
  3. Sep 18, 2012 #2

    SammyS

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    What are the solutions to the system of equations:

    x7 + y7 =1

    x + y = 1

    ?
     
  4. Sep 19, 2012 #3
    I think I'm very close to solving it. I substituted y= (1-x) in the first equation and expanded. Then I factored that. However I don't know how to factor the last term.

    x4-2x3+3x2-2x+1

    WolframAplha say it can be factored in to (x2-x+1)2. I just need to figure out how to factor that in to this and then I'm done (I think). Any ideas? Thanks.
     
  5. Sep 20, 2012 #4

    Mentallic

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    How did you get to that expression?
    Expanding (1-x)7 allows us to cancel the 1 and -x7 terms, so the highest power of x is 6, then after dividing through by x it should be a max power of 5.
     
  6. Sep 20, 2012 #5
    Exactly. After expanding and canceling out we have:

    7x6 - 21x5 + 35x4 - 35x3 + 21x2 - 7x.

    First I divided by (7x). Then I realized (x-1) is a factor (since x=1 is a zero). After long division I have:

    (7x)(x-1)(x4 - 2x3 + 3x2 - 2x + 1)
     
  7. Sep 20, 2012 #6

    Mentallic

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    Ahh ok so you already factored out the x=1 factor.

    So it's a quartic and hence it must have 4 complex roots. But since all the coefficients are real, the complex roots must come in complex conjugate pairs, and when you expand out [itex](x-\alpha)(x-\beta)[/itex] where [itex]\alpha, \beta \inℂ[/itex] it must be equal to a quadratic with real coefficients.

    So, with this we can deduce that the quartic must be able to be factorized into

    [tex](x^2+ax\pm 1)(x^2+bx\pm 1)[/tex]

    And expanding that, then equating coefficients we can deduce a=b=-1 and we need to take the positive of the [itex]/pm[/itex] operator.
     
  8. Sep 20, 2012 #7

    Ray Vickson

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    One way to proceed is to look for the intersection of the two graphs y = 1 - x and y = (1-x^7)^(1/7). For the latter: look at the graph y vs x for x^n + y^n = 1. When n = 1 you get the line x+y=1. When n = 2 you get the circle x^2 + y^2 = 1, a circle of radius 1 passing through the points (1,0) and (0,1). What happens if n > 2? Well, any point (x,y) in the interior of the first quadrant and on the circle x^2 + y^2 = 1 must lie to the left and below the curve y = f(x) for x^n + y^n = 1. This is because such a point on the circle has 0 < x < 1 and 0 < y < 1, so x^n < x^2 and y^n < y^2, hence x^n + y^n < 1. That means we need to increase x and/or y to bring the quantity x^n + y^n up to 1. In other words, for n > 2 the graph is outside the circle except at the ends (1,0) and (0,1). That means that the intersection of the graph with x + y = 1 is easy to ascertain.

    RGV
     
  9. Sep 20, 2012 #8

    SammyS

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    Split the 3x2 into x2 + 2x2 .

    x4-2x3+3x2-2x+1
    =x4-2x3+x2 + 2x2-2x+1

    =(x2-x)2 + 2(x2-x) + 1

    ...​
     
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