Results 1 to 3 of 3

Thread: Find all pairs of positive integers whose squares differ by 400?

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    19

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    $\displaystyle
    m^2 - n^2 = 400\quad m,n \in Z^ + $


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

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, Rinnie!

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

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

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

    Solve the system: .$\displaystyle \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 $\displaystyle x\text{ and }y$ are integers, $\displaystyle P\text{ and }Q$ have the same parity;
    . . both even or both odd.

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


    Then we have:

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Positive integers that are differences of squares
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Jul 2nd 2011, 07:11 PM
  2. Sums of squares of positive integers prime to n
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Nov 23rd 2009, 11:15 PM
  3. Find all positive integers such that...?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jul 20th 2008, 11:20 AM
  4. Replies: 4
    Last Post: Mar 28th 2007, 05:10 PM
  5. Replies: 2
    Last Post: Jul 5th 2005, 06:20 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum