Results 1 to 8 of 8

Math Help - Sum of 2 squares

  1. #1
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468

    Sum of 2 squares

    Sorry before I tell you the question here was the question that led to this question.

    Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2

    Done that successfully.

    Using this results, write 500050 as the sum of 2 square numbers.

    I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Mukilab View Post
    Sorry before I tell you the question here was the question that led to this question.

    Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2

    Done that successfully.

    Using this results, write 500050 as the sum of 2 square numbers.

    I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
    Try m=7.

    50 is very close to 49 and stands out that way..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, Mukilab!

    Show that: . m+n)^2+(mn-1)^2" alt=" (m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2" />

    Done that successfully. . Good!
    Using this results, write 500,050 as the sum of 2 square numbers.

    Note that: . 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)


    \text{Let }m = 7,\;n = 100 \text{ in the formula:}

    . . (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie hungthinh92's Avatar
    Joined
    Jun 2010
    From
    Virginia, US
    Posts
    23
    m=7 is a really nice guess
    OK. 500050=10001*50
    and 10001=10000+1
    and 50=49+1
    Apply in your equation:  (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
    Then....
    Last edited by hungthinh92; June 8th 2010 at 01:05 PM. Reason: too late...:(
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    That would make n a decial (10sqrt10)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by Soroban View Post
    Hello, Mukilab!




    Note that: . 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)


    \text{Let }m = 7,\;n = 100 \text{ in the formula:}

    . . (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2


    Thanks Soroban!

    Sorry, only saw the first post at the start

    Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Mukilab View Post
    That would make n a decial (10sqrt10)
    Maybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Mukilab View Post
    Thanks Soroban!

    Sorry, only saw the first post at the start

    Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?

    Thanks again
    Another way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sum of squares
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 16th 2010, 06:25 PM
  2. Magic squares of Squares
    Posted in the Math Puzzles Forum
    Replies: 5
    Last Post: September 22nd 2010, 09:58 AM
  3. Squares
    Posted in the Math Puzzles Forum
    Replies: 6
    Last Post: December 2nd 2009, 06:55 PM
  4. Replies: 4
    Last Post: November 13th 2009, 05:12 PM
  5. sum of squares
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 7th 2007, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum