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[WSC18] Proof of a Diophantine Equation that outputs Fibonacci Numbers

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Introduction

In Stephen Wolfram's book A New Kind of Science, he discovered a function that he conjectures to have some very special properties. He conjectures that for only positive inputs, all of the positive outputs produce Fibonacci Numbers, where each successive element greater than or equal to 3 is the sum of its previous elements. The function which he discovered follows as such: (2-(x^2-y^2-xy)^2)x The first Fibonacci Numbers up to 1000, produced  by the Diophantine Equation

This post will go through the proof that that specific Diophantine Equation produces Fibonacci Numbers for positive inputs and outputs. The goal of this post is to show a method that can be used to prove Dr Wolfram's Conjecture.

Proof

If we analyze the x^2-y^2-xy part, if this is an integer K such that K^2<2, then we will have discovered what x and y values satisfy the initial conditions stated above. The only integers squared that is less than 2 are 1, 0, and -1 (We can have a negative value because it will be squared).

So, using Mathematica, we can find the x and y values satisfying the conditions: First setting x^2-y^2-xy equal to 0 and solving: Quadratic = 0

Now setting x^2-y^2-xy equal to 1 and solving: enter image description here

Now setting x^2-y^2-xy equal to -1 and solving: enter image description here

We know that the first 100 Fibonacci Numbers are {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 10610209857723, 17167680177565, 27777890035288, 44945570212853, 72723460248141, 117669030460994, 190392490709135, 308061521170129, 498454011879264, 806515533049393, 1304969544928657, 2111485077978050, 3416454622906707, 5527939700884757, 8944394323791464, 14472334024676221, 23416728348467685, 37889062373143906, 61305790721611591, 99194853094755497, 160500643816367088, 259695496911122585, 420196140727489673, 679891637638612258, 1100087778366101931, 1779979416004714189, 2880067194370816120, 4660046610375530309, 7540113804746346429, 12200160415121876738, 19740274219868223167, 31940434634990099905, 51680708854858323072, 83621143489848422977, 135301852344706746049, 218922995834555169026, 354224848179261915075}

Now if we solve the first equation for the first 50 outputs we get: enter image description here It can be seen that the equation produces every sixth Fibonacci number starting from the third element so using Binet's formula to produce every nth Fibonacci, by making these two equations equal each other, it can be proven that the equation (left) produces only Fibonacci numbers. enter image description here We must show that the above equation is true, and takes on the form, enter image description here , and that enter image description here Setting enter image description here, we can do the following:

enter image description here

Following that logic, the proof can be extended to the other 11 equations as shown below:

enter image description here enter image description here

The code for the remaining proofs has been attached to this post.

Attachments:
POSTED BY: Suhaas Kataria
Answer
9 days ago

Great job! You might be interested in a simpler proof. One approach is to use experimental mathematics by playing around and discovering statements that seem true. If you accumulate enough then you'll find the right ones that can form lemmas to build a proof. Another approach is to look at the note in the book. Wolfram says that x^2-x*y-y^2=+-1 exactly when x and y are consecutive Fibonacci numbers. You could prove that lemma. Take a look at this expression for starters.

Simplify[(a + b)^2 - (a + b)*b - b^2]
POSTED BY: Todd Rowland
Answer
6 days ago

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