If an Even number could be expressed in the form a² + b² . And if there exits two other numbers m,n such that

a² + b² = m² + n²

then , my question is

is there any relation between (a,b) and (m,n) apart from a² + b² = m² + n² ??

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- Apr 6th 2011, 11:59 PMssnmanikantasum of two squares?
If an Even number could be expressed in the form a² + b² . And if there exits two other numbers m,n such that

a² + b² = m² + n²

then , my question is

is there any relation between (a,b) and (m,n) apart from a² + b² = m² + n² ?? - Apr 7th 2011, 05:39 AMOpalg
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, $\displaystyle 65 = 8^2+1^2 = 7^2+4^2$, and $\displaystyle 65 = 5\times13$, 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.

- Apr 7th 2011, 07:23 AMSoroban
Hello, ssnmanikanta!

Quote:

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

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

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

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

If a number is of the form $\displaystyle N \:=\:(a^2+b^2)(c^2+d^2)$, the product of two sums of squares,

. . then $\displaystyle \,N$ can be expressed as the sum of two squares in two ways.

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

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

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

$\displaystyle \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 \\

(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}$