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Solutions to a simple equation

  1. Dec 1, 2004 #1
    Are there integer (nonzero) solutions to this equation?

    [tex]a^2 + b^2 = c^2 + d^2 [/tex]
  2. jcsd
  3. Dec 1, 2004 #2
    yes, a=b=c=d=1, a=b=c=d=2, a=b=c=d=-1, etc.

    Or if you like, a=1=c, b=-1=d. Or whatever combination you can think of.
  4. Dec 1, 2004 #3
    Now that we've got that taken care of, how about solutions where a,b,c,d are unique?

    EDIT: Nevermind, I found one [tex] 6^2+7^2 = 2^2+9^2[/tex]
    Last edited: Dec 1, 2004
  5. Dec 1, 2004 #4


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    let (a/b, c/d) and (x/y, z/w) be any two "rational points" on the unit circle, i.e. points both of whose coordinates are rational numbers.

    then (a/b)^2 + (c/d)^2 = 1 = (x/y)^2 + (z/w)^2, and thus after

    multiplying out the denominators we get y^2w^2[a^2 + c^2] = b^2d^2[x^2 + z^2]

    which solves your problem.

    now there are infinitely many rational points on the unit circle since you can
    parametrize" the unit circle by projection from the north pole. i.e. join the point (0,1) to any rational point on the x axis. the line will meet the unit circle in a rational point. (a line with rational slope which meets the unit circle at one rational point, namely (0,1), also meets it at a second rational point.)

    try this one: solve a^3 + b^3 = x^3 + y^3, two different ways.
    Last edited: Dec 1, 2004
  6. Dec 1, 2004 #5
    A very insightful explanation, much appreciated.

    When I was thinking about my problem, I suddenly remembered that Ramanujan knew that [tex] 1729 = 9^3+10^3 = 1^3+12^3[/tex], so I figured mine had solutions too.
  7. Dec 1, 2004 #6


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    yes! that's the famous example i had in mind, supposedly provided by ramanujan to hardy in a taxicab, upon seeing the cabdrivers license number: 1729.

    question: does a^3 + b^3 = x^3 + y^3, have an infinite number of solutions?
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