How can I find the value n satisfying the equation?

Question

I would like to find the value n satisfying following equation . How can I do it in Mathematica?

15411 == 123*Fibonacci[n] + 31*Fibonacci[n + 1]

FindInstance[ 15411 == (123*Fibonacci[n] + 31*Fibonacci[n + 1]), n, Reals] — george2079, Commented Apr 5, 2018 at 14:03
ah, v10.1 yields 11.. (oddly it does not find the result if you make domain Integers ) — george2079, Commented Apr 5, 2018 at 15:19
Try FindInstance[ 15411 == (123*Fibonacci[n] + 31*Fibonacci[n + 1]) && n > 0, n, Reals], which yields the result $n=11$ (MMA v. 11.2.0 otherwise yields a negative root near $-12.7544$). — user15994, Commented Apr 5, 2018 at 15:20
Using a finite interval makes it return 11: FindInstance[{15411 == (123*Fibonacci[n] + 31*Fibonacci[n + 1]), 0 < n < 200000, n ∈ Integers}, n] — Coolwater, Commented Apr 5, 2018 at 15:47
Fibonacci is defined on reals other than integers, so could just use FindRoot. In[6]:= FindRoot[15411 == 123*Fibonacci[n] + 31*Fibonacci[n + 1],{n,9}] Out[6]= {n -> 11.} — Daniel Lichtblau, Commented Apr 5, 2018 at 20:58

Bob Hanlon · Accepted Answer · 2018-04-05 17:43:59Z

1

You can use either Solve or Reduce if you specify an range for n which contains the solution. The range does not have to be tight.

eqn = 15411 == 123*Fibonacci[n] + 31*Fibonacci[n + 1];

sol1 = Solve[{eqn, -50 < n < 50}, n, Integers]

(* {{n -> 11}} *)

Verifying,

eqn /. sol1[[1]]

(* True *)

Or

sol2 = Reduce[{eqn, -50 < n < 50}, n, Integers] // ToRules

(* {n -> 11} *)

answered Apr 5, 2018 at 17:43

Bob Hanlon

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1

$\begingroup$ or even .. (Pick[#, (eqn /. n -> # ) & /@ #] &@Range[-50, 50] ) $\endgroup$
– george2079
Commented Apr 5, 2018 at 17:56

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