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Thread: Perfect squares

  1. #1
    Newbie darknight's Avatar
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    Perfect squares

    Determine all positive integers n such that n+100 and n+168 are both perfect squares:
    Here' my attempt:
    n+168 should be less than 35sqr
    Since, 35sqr - 34sqr = 69 which is greater than 68, the difference between n+100 and n+168

    Beyond that, I've got no clue.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by darknight View Post
    Determine all positive integers n such that n+100 and n+168 are both perfect squares:
    Here' my attempt:
    n+168 should be less than 35sqr
    Since, 35sqr - 34sqr = 69 which is greater than 68, the difference between n+100 and n+168

    Beyond that, I've got no clue.
    Hi darknight,

    I would proceed by finding all qualifying perfect squares, then from them finding n.

    $\displaystyle n+100=a^2$

    $\displaystyle n+168=b^2$

    So

    $\displaystyle a^2-100=b^2-168$

    $\displaystyle b^2-a^2=68$

    $\displaystyle (b-a)(b+a)=68$

    We know that $\displaystyle (b-a)$ and $\displaystyle (b+a)$ are integers, so this problem is now equivalent to factoring 68 into two divisors, $\displaystyle 68=d_1d_2$.

    For any pair $\displaystyle (d_1,d_2)$, we can solve

    $\displaystyle (b-a) = d_1$

    $\displaystyle (b+a) = d_2$

    $\displaystyle \Rightarrow 2b = d_1+d_2$

    First, try $\displaystyle 68=1\cdot68$

    $\displaystyle \Rightarrow 2b = 69$

    No integer solutions.

    Next try $\displaystyle 68=2\cdot34$

    $\displaystyle \Rightarrow 2b = 36$

    $\displaystyle b = 18$

    $\displaystyle a = 16$

    $\displaystyle n = a^2 - 100 = 156$

    Next try $\displaystyle 68=4\cdot17$

    Again there are no integer solutions.

    So the only possible n is $\displaystyle \boxed{156}$.
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  3. #3
    Newbie darknight's Avatar
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    Ah well, went till (b-a)(b+a) = 68 and then hit the brick wall..
    Thanks
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