Given that each of x and y can be expressed as a sum of square of 2 integers, i.e.
x = a2 + b2
y = c2 + d2
where a, b, c and d are integers.
Show that xy can also be expressed as a sum of square of 2 integers.
Product of sum of squares
2009-08-05 5:33 pm
回答 (3)
2009-08-05 8:43 pm
✔ 最佳答案
Let m=a+bi,n=c+di, then mn=(ac-bd)+(ac+bd)i|m|=√(a^2+b^2), |n|=√(c^2+d^2)
|mn|=√[(ac-bd)^2+(ac+bd)^2]
since |m||n|=|mn|
√(a^2+b^2)√(c^2+d^2)=√[(ac-bd)^2+(ac+bd)^2]
Square both sides,
(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ac+bd)^2
xy=(ac-bd)^2+(ac+bd)^2
So xy can also be expressed as a sum of square of 2 integers.
2009-08-10 8:29 pm
xy
= (a² + b²)(c² + d²)
= a²c² + a²d² + b²c² + b²d²
= a²c² + b²d² + a²d² + b²c²
= a²c² + b²d² - 2acbd + 2acbd + a²d² + b²c²
= (a²c² - 2acbd + b²d²) + (a²d² + 2acbd + b²c²)
= (ac - bd)² + (ad + bc)²
As a, b, c and d are integers,
ac - bd and ad + bc are also integers.
So, xy can also be expressed as a sum of square of 2 integers.
= (a² + b²)(c² + d²)
= a²c² + a²d² + b²c² + b²d²
= a²c² + b²d² + a²d² + b²c²
= a²c² + b²d² - 2acbd + 2acbd + a²d² + b²c²
= (a²c² - 2acbd + b²d²) + (a²d² + 2acbd + b²c²)
= (ac - bd)² + (ad + bc)²
As a, b, c and d are integers,
ac - bd and ad + bc are also integers.
So, xy can also be expressed as a sum of square of 2 integers.
2009-08-05 10:03 pm
Since √(a^2+b^2)√(c^2+d^2)=√[(ac-bd)^2+(ac+bd)^2],
we have |m||n|=|mn|
we have |m||n|=|mn|
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