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Math Help - Find all pairs of positive integers whose squares differ by 400?

  1. #1
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    Find all pairs of positive integers whose squares differ by 400?

    I have pairs, but I need to find an organzied method for finding them?
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
m^2  - n^2  = 400\quad m,n \in Z^ +


    \left( {m - n} \right)\left( {m + n} \right) = 400

    thus m-n, or m+n must equal one of the divisors of 400, to satisfy the equation, and so that m,n will be positive integers.

    for example, 10 is a divisor of 400:

    m - n = 10
    m + n = 40

    -> m = 25, n = 15
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  3. #3
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    Hello, Rinnie!

    Find all pairs of positive integers whose squares differ by 400.

    Let x,y be positive integers such that: . x^2-y^2 \:=\:400

    Then we have: . (x+y)(x-y) \:=\:P\cdot Q ... where P\cdot Q \:=\:400

    Solve the system: . \begin{array}{ccc}x + y &=& P \\ x-y&=& Q\end{array}\quad\Rightarrow\quad\begin{Bmatrix}x &=&\frac{P+Q}{2} \\ \\[-4mm] y &=&\frac{P-Q}{2} \end{Bmatrix}

    Since x\text{ and }y are integers, P\text{ and }Q have the same parity;
    . . both even or both odd.

    The only ways to factor 400 into two factors with the same parity are:
    . . (200, 2),\;\;(100,4,),\;\;(50,8),\;\;(40,10),\;\;(20,20)


    Then we have:

    . . \begin{array}{c|cc}<br />
(P,Q) & (x,y) \\ \hline (200,2) & (101,99) \\ (100,4) & (52,48) \\ (50,8) & (29,21) \\ (40,10) & (25,15) \\ (20,20) & {\color{red}\rlap{/////}}(20,0) & {\color{red}\text{not positive}}\end{array}


    Therefore: . 101^2-99^2 \:=\:52^2-48^2 \:=\:29^2-21^2 \:=\:25^2-15^2 \:=\:400

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