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^2v^2= uv1. [1] (Those F_(n) for which u^2v^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^2v^2+1) = 4u^22v^2+2 = 2(2UV+1)
b=d*h: v(2u+v) =v^2+2uv = 2u^2v^2+2 = 2UV+2
c=f^2+g^2: u^2+(u+v)^2 = 2u^2+2uv+v^2 = 4u^2v^2+2 = 4UV+2
a^2+b^2 = (4U2V+2)^2+(2UV+2)^2 = 1/5(10U5V+6)^2+4/5, by completing the square, so (10U5V+6)^2= 5(4UV+2)^24.
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^24 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_(n1))^2 = (F_(2n1)) (Source: Cassini's identity). Hence c= (F_(2g1)).
While this already looks like enough, here is a direct algebraic proof:
b, a ,c, share the common element, UV+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^24Ux+4U4x^2 = 0.
Slice and dice:
4U^2+4U+1 = 4x^24Ux+1 Rearranging, adding 1 to both sides
(2U+1)^24x^2=4Ux+1 Rearranging
(2U+1)^24x^2=4U(UV+1)+1 Substituting back on RHS
(2U+1)^24x^2=4U^2+4U+14UV Expanding
(2U+1)^24x^2=(2U+1)^24UV Rearranging
4(UV+1)^2=4UV Cancelling like terms
(UV+1)^2=UV Dividing by 4.
But UV+1 = u^2v^2+1, while UV =u^2v^2, so we can take their square roots:
u^2v^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

Posted by broll
on 20140626 03:32:23 