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Finding an integer n that satisfies the equation 5 divides 2^n+1?

I am looking for a way to find an integer that satisfies the statement 5 is divisible by 2^n+1. The answer is 2 and I have tried Solve, Reduce, and FindInstance and not solved the equation.

Solve[Divisible[5,2^n+1]==True,n]
Reduce[Divisible[5,2^n+1]==True,n]
FindInstance[Divisible[5,2^n+1]==True,n]
Solve[Mod[5,2^n+1]==0,n]
Reduce[Mod[5,2^n+1]==0,n]
FindInstance[Mod[5,2^n+1]==0,n]

Is there a way to solve for the answer of 2?

POSTED BY: Peter Burbery
3 Replies

In fact every number

z[m_] := 2^(2 + 4 m ) + 1 

for m element Integers is divided by 5 without rest (proof by induction).

POSTED BY: Hans Dolhaine

And some more solutions

2^# + 1 & /@ {2, 6, 10, 14}
POSTED BY: Hans Dolhaine

This way you also get a negative answer:

In[83]:= Solve[Mod[5, 2^n + 1] == 0, n, Integers]

Out[83]= {{n -> -2}, {n -> 2}}
POSTED BY: Gianluca Gorni
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