-1

I have discovered an identity which generalizes Vajda's identity concerning Fibonacci Numbers. The identity states that: $$F_{n+i+x-z}F_{n+j+y+z}-F_{n+x+y-k}F_{n+i+j+k}=(-1)^{n+x+y-k}F_{i+k-y-z}F_{j+k-x+z}$$

Note that Vajda's identity states that: $$F_{n+i}F_{n+j}-F_{n}F_{n+i+j}=(-1)^{n}F_{i}F_{j}$$

So, I want to basically ask if this kind of result is publishable.

Shuaib Lateef
  • 439
  • To what extent have you verified this? Are there any requirements that all of the subscripts are positive, for example? I've tried this numerically with random positive $n,i,j,k,x,y,z$ and it does not work. – Cye Waldman Apr 03 '21 at 22:39
  • There was a slight mistake and I have corrected it. Sorry for the inconvenience. – Shuaib Lateef Apr 04 '21 at 08:44
  • That solved the problem. The identity seems to be correct. Now, in answer to your question about originality, I suspect that it is. But you should check Vajda's book Fibonacci and Lucas Numbers, and the Golden Section, Dover, 1989 and Ron Knott's web pages at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fib.html as well. Also, it can't hurt to submit it to a Journal and see what happens. I'm not sure if The Fibonacci Quarterly is interested in something like this. If you do want to publish, you'll have to show a derivation. – Cye Waldman Apr 04 '21 at 15:38
  • I actually have a derivation. I just wanted to confirm if I won't be preparing the result for publication in vain. Thanks for this comment. – Shuaib Lateef Apr 04 '21 at 16:59
  • The link below contains the paper I have prepared for publication on the identity. Is the proof correct and besides, is the paper written well enough to be published? vixra.org/abs/2102.0163 – Shuaib Lateef Apr 11 '21 at 21:07
  • 2
    asked and answered at https://mathoverflow.net/q/396786/11260 – Carlo Beenakker Jul 04 '21 at 17:42

1 Answers1

0

Proof

$$F_{n+i+x-z}F_{n+j+y+z}-F_{n+x+y-k}F_{n+i+j+k}=(-1)^{n+x+y-k}F_{i+k-y-z}F_{j+k-x+z} \tag{1}$$

We know from Binet's formula for generating nth Fibonacci number that
if $$\phi = \frac{1+\sqrt 5}{2}, \varphi = \frac{1-\sqrt 5}{2}$$ then

$$F_n = \frac{\phi^n - \varphi^n}{\sqrt 5}$$

where $F_n$ is the nth Fibonacci number\

From (1), let

$$\alpha = F_{n+i+x-z}F_{n+j+y+z} \tag{2}$$

$$\beta = F_{n+x+y-k}F_{n+i+j+k}\tag{3}$$

$$\gamma = (-1)^{n+x+y-k} F_{i+k-y-z}F_{j+k+z-x}\tag{4}$$

such that $$\alpha -\beta = \gamma \tag{5}$$

We can see that to prove (1), it suffices to show that (5) is true\

Also, let $$P_1=\frac{\phi^{2n+i+x+j+y}}{5},$$ $$P_2=\frac{\varphi^{2n+i+x+j+y}}{5},$$ $$P_3=\frac{\phi^{n+i+x-z}\varphi^{n+j+y+z}}{5},$$ $$P_4=\frac{\phi^{n+j+y+z}\varphi^{n+i+x-z}}{5},$$ $$P_5=\frac{\phi^{n+x+y-k}\varphi^{n+i+j+k}}{5},$$ $$P_6=\frac{\phi^{n+i+j+k}\varphi^{n+x+y-k}}{5}$$ $$P_7=\phi^{i+k-y-z}\varphi^{j+z+k-x},$$ $$P_8=\phi^{j+z+k-x}\varphi^{i+k-y-z},$$ $$P_9=\phi^{2k+i+j-x-y},$$ $$P_{10}=\varphi^{2k+i+j-x-y}$$ $$P_{11}= F_{i+k-y-z}F_{j+k+z-x}$$

From (2), we see that

$$\alpha = F_{n+i+x-z}F_{n+j+y+z}$$ $$\alpha = \left(\frac{\phi^{n+i+x-z}- \varphi^{n+i+x-z}}{\sqrt 5}\right)\left(\frac{\phi^{n+j+y+z}- \varphi^{n+j+y+z}}{\sqrt 5}\right)$$

$$\alpha = \frac{\phi^{2n+i+x+j+y}}{5}-\frac{\phi^{n+i+x-z}\varphi^{n+j+y+z}}{5}-\frac{\phi^{n+j+y+z}\varphi^{n+i+x-z}}{5}+\frac{\varphi^{2n+i+x+j+y}}{5}$$

$$\alpha = P_1 - P_3 - P_4 + P_2 \tag{6}$$

From (3), we see that

$$\beta = F_{n+x+y-k}F_{n+i+j+k}$$ $$\beta = \left(\frac{\phi^{n+x+y-k}- \varphi^{n+x+y-k}}{\sqrt 5}\right)\left(\frac{\phi^{n+i+j+k}- \varphi^{n+i+j+k}}{\sqrt 5}\right)$$

$$\beta =\frac{\phi^{2n+i+x+j+y}}{5} - \frac{\phi^{n+x+y-k}\varphi^{n+i+j+k}}{5}-\frac{\phi^{n+i+j+k}\varphi^{n+x+y-k}}{5}+ \frac{\varphi^{2n+i+x+j+y}}{5}$$

$$\beta = P_1 - P_5 - P_6 + P_2 \tag{7}$$

Deducting (7) from (6) gives $$\alpha - \beta = (P_5 + P_6) -(P_3 + P_4)\tag{8}$$

From (8), let $$V_1 = -(P_3 + P_4)$$ $$V_2 = (P_5 + P_6)$$

then

$$V_1 = -\left(\frac{\phi^{n+i+x-z}\varphi^{n+j+y+z}}{5} + \frac{\phi^{n+j+y+z}\varphi^{n+i+x-z}}{5}\right)$$ $$V_1 = -\frac{(\phi\varphi)^{n+x+y-k}}{ (\phi\varphi)^{n+x+y-k} }\left(\frac{\phi^{n+i+x-z}\varphi^{n+j+y+z}}{5} + \frac{\phi^{n+j+y+z}\varphi^{n+i+x-z}}{5}\right)$$ $$V_1 = - \frac{1}{5}((\phi\varphi)^{n+x+y-k})\left(\frac{\phi^{n+i+x-z}}{\phi^{n+x+y-k}} \frac{\varphi^{n+j+y+z}}{\varphi^{n+x+y-k}}+ \frac{\phi^{n+j+y+z}}{\phi^{n+x+y-k}} \frac{\varphi^{n+i+x-z}}{\varphi^{n+x+y-k}}\right)$$

$$V_1 =-\frac{1}{5} ((\phi\varphi)^{n+x+y-k})(\phi^{i+k-y-z}\varphi^{j+z+k-x} + \phi^{j+z+k-x}\varphi^{i+k-y-z})$$ Note that $$\phi\varphi = -1$$ $$V_1 =-\frac{1}{5} ((-1)^{n+x+y-k})(P_7 + P_8)$$

Also,

$$V_2 = \left(\frac{\phi^{n+x+y-k}\varphi^{n+i+j+k}}{5} + \frac{\phi^{n+i+j+k}\varphi^{n+x+y-k}}{5}\right)$$ $$V_2 = \frac{ (\phi\varphi)^{n+x+y-k} }{ (\phi\varphi)^{n+x+y-k} }\left(\frac{\phi^{n+x+y-k}\varphi^{n+i+j+k}}{5} + \frac{\phi^{n+i+j+k}\varphi^{n+x+y-k}}{5}\right)$$ $$V_2 = \frac{1}{5} ((\phi\varphi)^{n+x+y-k})\left(\frac{\phi^{n+x+y-k}}{\phi^{n+x+y-k}} \frac{\varphi^{n+i+j+k}}{\varphi^{n+x+y-k}}+ \frac{\phi^{n+i+j+k}}{\phi^{n+x+y-k}} \frac{\varphi^{n+x+y-k}}{\varphi^{n+x+y-k}}\right)$$

$$V_2 = \frac{1}{5}((\phi\varphi)^{n+x+y-k})(\phi^{0} \varphi^{2k+i+j-x-y} + \phi^{2k+i+j-x-y}\varphi^{0})$$

$$V_2 = \frac{1}{5}((\phi\varphi)^{n+x+y-k})(\phi^{2k+i+j-x-y} + \varphi^{2k+i+j-x-y})$$

Note that $$\phi\varphi = -1$$

$$V_2 = \frac{1}{5}((-1)^{n+x+y-k})(P_9 + P_{10})$$

Now, we see from (8) that

$$\alpha - \beta =(P_5 + P_6) -(P_3 + P_4)$$ $$\alpha - \beta = V_1 + V_2$$

$$\alpha - \beta = \frac{1}{5} ((-1)^{n+x+y-k})(P_9 -P_7 - P_8 +P_{10})\tag{9}$$

From (4), we see that

$$\gamma = (-1)^{n+x+y-k} F_{i+k-y-z}F_{j+k+z-x}$$

$$\gamma = (-1)^{n+x+y-k}(P_{11})\tag{10}$$

But

$$P_{11} = F_{i+k-y-z}F_{j+k+z-x}$$

$$P_{11} = \left(\frac{\phi^{i+k-y-z}- \varphi^{i+k-y-z}}{\sqrt 5}\right)\left(\frac{\phi^{j+k+z-x}- \varphi^{j+k+z-x}}{\sqrt 5}\right)$$

$$P_{11} = \frac{1}{5}(\phi^{i+k-y-z}- \varphi^{i+k-y-z})(\phi^{j+k+z-x}- \varphi^{j+k+z-x})$$ $$P_{11} = \frac{1}{5}(\phi^{2k+i+j-x-y} - \phi^{i+k-y-z}\varphi^{j+z+k-x} - \phi^{j+z+k-x}\varphi^{i+k-y-z} +\varphi^{2k+i+j-x-y})$$

$$P_{11} = \frac{1}{5}(P_9 - P_7 - P_8 + P_{10})\tag{11}$$

So, putting (11) in (10) gives

$$\gamma = \frac{1}{5}(-1)^{n+x+y-k}(P_9 - P_7 - P_8 + P_{10})\tag{12}$$

Since (12) equals (9) then, (5) is true which completes the proof.\

Shuaib Lateef
  • 439