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Math Help - sum of two squares?

  1. #1
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    sum of two squares?

    If an Even number could be expressed in the form a + b . And if there exits two other numbers m,n such that
    a + b = m + n

    then , my question is

    is there any relation between (a,b) and (m,n) apart from a + b = m + n ??
    Last edited by mr fantastic; April 7th 2011 at 05:11 AM. Reason: Deleted excess ? in title.
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  2. #2
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    Quote Originally Posted by ssnmanikanta View Post
    If an Even number could be expressed in the form a + b . And if there exits two other numbers m,n such that
    a + b = m + n

    then , my question is

    is there any relation between (a,b) and (m,n) apart from a + b = m + n ??
    The condition for an integer x to be a sum of two squares in more than one way is that x should have at least two prime factors of the form 4k+1. For example, 65 = 8^2+1^2 = 7^2+4^2, and 65 = 5\times13, with both 5 and 13 being a multiple of 4 plus 1. But there is no particular relation between the pairs of numbers (1,8) and (4,7) occurring in those decompositions.
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  3. #3
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    Hello, ssnmanikanta!

    \text{If an even number could be expressed in the form }a^2 + b^2\text{. and if}

    \text{there exits two other numbers }m,n\text{ such that: }\,a^2 + b^2 \:=\: m^2 + n^2,

    \text{then is there any relation between }(a,b)\text{ and }(m,n)\,?

    There is an interesting theorem that may or may not help . . .


    If a number is of the form N \:=\:(a^2+b^2)(c^2+d^2), the product of two sums of squares,
    . . then \,N can be expressed as the sum of two squares in two ways.

    . . N \;=\;(a^2+b^2)(c^2+d^2) \;=\;\begin{Bmatrix}(ac + bd)^2 + (ad-bc)^2 \\ (ac-bd)^2 + (ad+bc)^2 \end{Bmatrix}


    \text{Example: }\;a = 5,\;b = 2,\;c = 4,\;d = 3

    \text{Then: }\:N \;=\;(5^2+2^2)(4^2+3^2) \:=\:29\cdot25 \;=\;725

    \text{Hence: }\;\begin{array}{ccccccccc}(5\!\cdot\!4 + 2\!\cdot\!3)^2 + (5\!\cdot\!3 - 2\!\cdot\!4)^2 &=& (20 + 6)^2 + (15 - 8)^2 &=& 26^2 + 7^2 &=& 725 \\<br />
(5\!\cdot\!4 - 2\!\cdot\!3)^2 + (5\!\cdot\!3 + 2\!\cdot\!4)^2 &=& (20-6)^2 + (15+8)^2 &=&  14^2 + 23^2 &=& 725 \end{array}

    Last edited by Soroban; April 7th 2011 at 07:38 AM.
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