Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \[ M_2 = \left(\f...
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Proof of the theorem regarding Fibonacci prime numbers

Hello everyone. Today we will prove the theorem regarding Fibonacci prime numbers. We will use the method of contradiction as a proof method. 

Theorem: Let FnF_n be the nth Fibonacci number and let FnF_n be a prime number. Then nn is also a prime number, except in the case of F4=3F_4 = 3 . 

Proof: 

For the case when n=2n = 2 we have that F2=1F_2 = 1 , and as we know the number 11 is neither simple nor composite, so this case does not refute the truth of the theorem. 

For the case when n=3n = 3 we have that F3=2F_3 = 2 , so this case is in accordance with the statement. 

Now, suppose that for n>4n> 4 , FnF_n is a prime number and that n=rsn = rs for some natural numbers r,sr, s which are greater than 11 , that is, that nn is a composite number. 

Since n>4n> 4   at least one of the numbers rr and ss is greater than 22 . Then, according to the divisibility theorem of Fibonacci numbers which reads: ∀m,n∈Z>2m∣n⇔Fm∣Fn\forall m, n \in \mathbb {Z} _ {> 2} \quad m \mid n \Leftrightarrow F_m \mid F_n we have that at least one of the expressions Fr∣FnF_r \mid F_n and Fs∣FnF_s \mid F_n is correct. Further, since at least one of the numbers rr and ss is greater than 22 , it follows that at least one of the Fibonacci numbers FrF_r and FsF_s is greater than 11 . So, we have shown that the number FnF_n has at least one divisor greater than 11 and less than FnF_n , i.e. we have shown that FnF_n is a composite number, thus we arrived at a contradiction. From this we conclude that nn must be a prime number. 

■\blacksquare  

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