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To answer the question of proving Fibonacci sequence is periodic mod 5 without using induction., I came across Mathematica to prove $$F_{n}\equiv F_{n+20}\pmod 5$$ for all $n \geq 2$

I defined:

$F[n] := \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$

and showed:

```
Simplify[f[n + 20] - f[n] - 10945 f[n] - 6765 f[n - 1]]
```

equals to 0. (I knew the fact that $F_{n+20} = 10946 F_{n}+ 6765 F_{n-1}$, but I wanted to reach a proof without induction)

Thus, for all $n \geq 2$ we have $$F_{n}\equiv F_{n+20}\pmod 5$$

The problem is that I can not get the intermediate steps of `Simplify`

function, even using the Trace commands. Could you please help?

If possible, I want to also show this in Wolfram Alpha. I don't know how to run several commands in Wolfram Alpha.

2

`Fibonacci[]`

is built-in. Also,`Simplify[]`

was never meant to show intermediate steps, as what it does under the hood is rather different from what a human would do. – J. M.'s ennui – 2015-07-13T18:03:42.763@Guesswhoitis. If you read my answer at math.SE, you see that I didn't want to calculate Fibonacci[n] with a reursive formula defined by induction. Because of this, I have used the closed form of Binet's formula. https://en.wikipedia.org/wiki/Jacques_Philippe_Marie_Binet

– Ho1 – 2015-07-13T18:09:10.1431

Yes, and

– J. M.'s ennui – 2015-07-13T18:11:17.700`FunctionExpand[Fibonacci[n]]`

will give you (a continued version of) Binet's formula. Whatever gave you the idea that`Fibonacci[]`

is unsuitable here? (FWIW, I am somewhat familiar with Binet.)Do you have any idea about how can I get the intermediate steps of simplifying? – Ho1 – 2015-07-13T18:13:03.660

2Again:

Mathematicadoes not work like that. It doesn't do things the way a human might, so not much point in having it "show its work". – J. M.'s ennui – 2015-07-13T18:15:55.260Do you mean that there is a possiblity that Mathematica does this simplifying numerically? It outputs

exact Zero. – Ho1 – 2015-07-13T18:18:13.530I am merely saying that a computer algebra system will in general not manipulate symbols in the same way a human might, nothing more. – J. M.'s ennui – 2015-07-13T18:20:57.707

So, I can say that my proof is mathematically correct. But I have a trouble to show others the proof. :-( Anyway, there should be a way, so I can see that how

it works, no matter how different it does this from a human. – Ho1 – 2015-07-13T18:23:36.320"The computer is a precision tool. It should not be used as a bludgeon or a substitute for thought." - F. S. Acton (See this.) – J. M.'s ennui – 2015-07-13T18:28:19.253

I don't want it to think instead of me, I want it to show the intermediate steps of simplifying an algebraic expression. – Ho1 – 2015-07-13T19:05:16.987