F2 Maths Hard Question

1)
(a) Prove that (m²-1)² + (2m)² = (m²+1)² is an identity.

LHS
= (m²-1)² + (2m)²
= m⁴ - 2m² + 1 + 4m²
= m⁴ + 2m² + 1
= (m²+1)²
= RHS

(b) Using (a), find positive integers a and b such that a² + b² = 65²

Consider m² + 1 = 65
m² = 64
m = 8 or -8 (rejected as the question requires positive integer)
m² - 1 = 63
2m = 16

Therefore, we have 63² + 16² = 65².

2013-09-15 15:01:11 補充：
hei, 你可以睇下, 這個是得出 Pythagorian triple的一個方法：

http://en.wikipedia.org/wiki/Pythagorean_triple

以本題為例，即是可以有一直角三角形，邊長為16, 63, 65。

LHS=(m^2-1)^2+(2m)^2
=m^4-m^2-m^2+1+4m^2
=m^4+2m^2+1
=(m^2+1)^2=RHS

Let m^2+1=65 a=m^2-1 b=2m
m=8 a=63 b=16

1(a) (m^2 - 1)^2 + (2m)^2

= m^4 - 2m^2 + 1 + 4m^2

= m^4 + 2m^2 + 1

=(m^2 + 1)^2

So, (m^2 - 1)^2 + (2m)^2 = (m^2 + 1)^2 is an identity

(b) Sub. m = 8, a = m^2 - 1 = 63 and b = 2m = 16