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Math Help - Question 14

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    Question 14

    We have two problems this week.
    First one comes from CaptainBlank.

    1)Show that if N has p divisors (including 1 and itself),
    that the product of all of these divisors is \sqrt{N^p}.

    2)A "Pythagorean Triangle" is a triangle with integer sides. For example, 3-4-5. Show that the area of a Pythagorean Triangle is never a square. (Fermat)
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    Quote Originally Posted by ThePerfectHacker View Post
    We have two problems this week.
    First one comes from CaptainBlank.

    ...

    2)A "Pythagorean Triangle" is a triangle with integer sides. For example, 3-4-5. Show that the area of a Pythagorean Triangle is never a square. (Fermat)
    Hello,

    to 2): (I don' know how to "whiten" the Latex formulae, therefore I post my reply completely visible)

    A right triangle with hypotenuse c and legs a and b has integer sides if
    a = u² - v², u > v and u, v in IN
    b = 2uv
    c = u² + v²

    (for instance u = 2, v = 1 will give the famous 3, 4, 5 triangle)

    The area of a right triangle is
    A=\frac{a \cdot b}{2}

    Plug in the above mentioned terms for a and b:

    A=\frac{(u^2+v^2) \cdot 2uv}{2}=(u^2-v^2) \cdot uv

    Now you have to prove that
    - u^2-v^2=(u+v)(u-v) is only a square if u=v or u=1 and v=0
    - u \cdot v is only a square if u = v
    - u^2-v^2=uv. Solve this equation for u and you'll get:
    u=\frac{v \pm v \cdot \sqrt{5}}{2} that means u is not in IN

    Thus the area can't be a square.

    EB

    Second thought:

    A=(u^2-v^2)uv=(u+v)(u-v)\cdot u \cdot v
    If this is a square there must be two pairs of equal factors which isn't possible under the given conditions. Thus A can't be a square.
    Last edited by earboth; January 12th 2007 at 09:45 AM.
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    Quote Originally Posted by earboth View Post
    Hello,

    to 2): (I don' know how to "whiten" the Latex formulae, therefore I post my reply completely visible)

    A right triangle with hypotenuse c and legs a and b has integer sides if
    a = uČ - vČ, u > v and u, v in IN
    b = 2uv
    c = uČ + vČ

    (for instance u = 2, v = 1 will give the famous 3, 4, 5 triangle)

    The area of a right triangle is
    A=\frac{a \cdot b}{2}

    Plug in the above mentioned terms for a and b:

    A=\frac{(u^2+v^2) \cdot 2uv}{2}=(u^2-v^2) \cdot uv

    Now you have to prove that
    - u^2-v^2=(u+v)(u-v) is only a square if u=v or u=1 and v=0
    - u \cdot v is only a square if u = v
    - u^2-v^2=uv. Solve this equation for u and you'll get:
    u=\frac{v \pm v \cdot \sqrt{5}}{2} that means u is not in IN

    Thus the area can't be a square.

    EB
    As defined by TPH, a "Pythagorean triangle" need only have integer length sides, it doesn't have to include a right angle.

    -Dan
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    Here's how you do it, earboth. Click on quote to see


    highlight between *earboth*
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    Quote Originally Posted by topsquark View Post
    As defined by TPH, a "Pythagorean triangle" need only have integer length sides, it doesn't have to include a right angle.
    That was a mistake by me. I assumed when I said "Pythagorean" it will be taken as right.
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    Quote Originally Posted by galactus View Post
    Here's how you do it, earboth. Click on quote to see


    highlight between *earboth*
    Hello, galactus,

    thanks a lot, but the problem I couldn't master is:

    Hi, here you have an equation written in Latex: y = x^2 and this isn't whitened

    EB
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    Quote Originally Posted by earboth View Post
    Hello, galactus,

    thanks a lot, but the problem I couldn't master is:

    Hi, here you have an equation written in Latex: y = x^2 and this isn't whitened

    EB
    It's a know problem (at least to some of us) that I have struggled with
    in the past, and not found a solution short of not using LaTeX in invisible
    text .

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    That was a mistake by me. I assumed when I said "Pythagorean" it will be taken as right.
    Well that makes the problem a lot easier and explains why I couldn't format a proof!

    -Dan
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    I was looking at earboth's approach but I did not understand it. For example if uv is a square only if u=v but that is not true, what happens when u=4,v=9?
    -----
    The "official" answer appeared in the note of a book owned by Fermat on Arithmetica by Diophantus.

    It would be easier if you saw these famous diophantine equations before, rather than work it out thyself. Fermat sates that,
    x^4+y^4=z^2
    x^4-y^4=z^2
    Has no solutions.
    (The first one leads to Fermat's Last Theorem n=4.)

    Here is the "offical" answer.
    Consider a Pythagorean triangle where z is hypotenuse. And x,y are the legs.
    Thus,
    x^2+y^2=z^2. (1)
    A=\frac{1}{2}xy.
    Assume,
    \frac{1}{2}xy=w^2.
    Then,
    2xy=4w^2=(2w)^2 (2)
    Add (2) and then subtract (2) from (1),
    (x+y)^2=z^2+(2w)^2 (3)
    (x-y)^2=z^2-(2w)^2 (4)
    Multiply (3) and (4) together,
    (x^2-y^2)^2=z^4-(2w)^4
    But this leads to one of Fermat's equations.... a contradiction.
    -----
    Now CaptainBlank's question.

    Let, N be a positive integer.
    And let,
    n_1,n_2,...,n_p
    Be all its factors (trivial and improper included).
    Let,
    M=n_1n_2....n_p (*)
    M=n_1n_2....n_p (**)
    Multiply them together,
    But before we do let me explain what happens.
    For each factor n_i in (*) we can find one in (**) n_j such that n_in_j=N.
    Thus,
    M^2=N\cdot N\cdot ... \cdot N=N^p
    Thus,
    M=\sqrt{N^p}.

    (In the special case when N is a square then p is odd but N is a square).
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