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Thread: Diophantine eq

  1. #1
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    Diophantine eq

    Thanks heaps but there is another one which I cant figure out. I have to prove that for each positive integer n there exist integers x,y,z > 1 so that:
    x^2+y^2=17z^(4n)

    :S

    How does induction work with 4 variables?
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  2. #2
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    No need for induction. Try to guess a solution which will be valid for all n (and dependent on n), using a single solution to x^2+y^2=17.

    (Sorry if this hint was too big)
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  3. #3
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    I think AdamC and I are in the same class ;-)

    The question asks us specifically to prove the proposition using simple induction. But, Unbeatable0, if I've taken your hint correctly, the values for x, y & z need to be greater than 1.
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  4. #4
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    Quote Originally Posted by AdamC View Post
    Thanks heaps but there is another one which I cant figure out. I have to prove that for each positive integer n there exist integers x,y,z > 1 so that:
    x^2+y^2=17z^(4n)

    :S

    How does induction work with 4 variables?


    Why do you worry about the variables? Worry about n...

    For n = 1:\,\,4^2+1^2=17\cdot 1^{4}...check

    Suppose for n:\,\,\exists x,y,z\,\,s.t.\,\,x^2+y^2=17z^{4n} , and we shall prove for n+1. But

    z^{4(n+1)}=z^4z^{4n} , so using the inductive hypothesis...

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Why do you worry about the variables? Worry about n...

    For n = 1:\,\,4^2+1^2=17\cdot 1^{4}...check
    Here, you should worry about the variables, because the question specifies integers x,y,z > 1. (But it's not hard to adapt the base case example to satisfy those conditions.)
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  6. #6
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    Quote Originally Posted by Opalg View Post
    Here, you should worry about the variables, because the question specifies integers x,y,z > 1. (But it's not hard to adapt the base case example to satisfy those conditions.)

    Good point. Completely missed that bigger than 1 thingy.

    Tonio
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