Let f=u,d=v, g=u+v, h=2u+v. Further, let u^2=U, v^2=V. Last, select u and v such that: u^2-v^2= uv-1. [1] (Those F_(n) for which u^2-v^2= uv+1 can be ignored, see Note).
These expansions and substitutions are then possible:
a=2*f*g: 2u(u+v)=2u^2+2(u^2-v^2+1) = 4u^2-2v^2+2 = 2(2U-V+1)
b=d*h: v(2u+v) =v^2+2uv = 2u^2-v^2+2 = 2U-V+2
c=f^2+g^2: u^2+(u+v)^2 = 2u^2+2uv+v^2 = 4u^2-v^2+2 = 4U-V+2
a^2+b^2 = (4U-2V+2)^2+(2U-V+2)^2 = 1/5(10U-5V+6)^2+4/5, by completing the square, so (10U-5V+6)^2= 5(4U-V+2)^2-4.
Note: 'The question may arise whether a positive integer x is a Fibonacci number. This is true if and only if one or both of 5x^2+4 or 5x^2-4 is a perfect square. (Source: Wiki).' f and g are consecutive Fibonacci numbers. The sum of the squares of two consecutive Fibonacci numbers is a Fibonacci number: (F_(n))^2+ (F_(n-1))^2 = (F_(2n-1)) (Source: Cassini's identity). Hence c= (F_(2g-1)).
While this already looks like enough, here is a direct algebraic proof:
b, a ,c, share the common element, U-V+1; call this x, giving U+x+1, 2U+2x, 3U+x+1.
Squaring and expanding:
5U^2+10Ux+2U+5x^2+2x+1 (a^2+b^2)
9U^2+6Ux+6U+x^2+2x+1 (c^2)
If c^2-(a^2+b^2) = 0, then 4U^2-4Ux+4U-4x^2 = 0.
Slice and dice:
4U^2+4U+1 = 4x^2-4Ux+1 Rearranging, adding 1 to both sides
(2U+1)^2-4x^2=4Ux+1 Rearranging
(2U+1)^2-4x^2=4U(U-V+1)+1 Substituting back on RHS
(2U+1)^2-4x^2=4U^2+4U+1-4UV Expanding
(2U+1)^2-4x^2=(2U+1)^2-4UV Rearranging
4(U-V+1)^2=4UV Cancelling like terms
(U-V+1)^2=UV Dividing by 4.
But U-V+1 = u^2-v^2+1, while UV =u^2v^2, so we can take their square roots:
u^2-v^2+1=uv. So a^2+b^2=c^2 if and only if this condition is met; see [1]
Coincidentally, a similar set of ideas was in play in Validity Vindication, because: U+x+1, 2U+2x, 3U+x+1 represents an integer RHT whose sides converge to 1,2,sqrt(5).
Edited on June 26, 2014, 4:25 am
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Posted by broll
on 2014-06-26 03:32:23 |
Solution
