Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x.Then for what value of p is the relation?

Question

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x.Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x
(a) 5
(b) 3
(c) 2
(d) 6

Answer 1

g(x) = g(x+1) + g(x-1)
g(x+1) = g(x) + g(x+2)
adding together we get
g(x) + g(x+1) = g(x) + g(x+1) + g(x-1) + g(x+2)
g(x+2) = -g(x-1)
g(x) = -g(x+3)
g(x+3) = -g(x+6)
so g(x) = g(x+6)
So it must be true for p = 6.
Consider the function defined on only the integers, with the sequence
.... 1 4 3 -1 -4 -3 1 4 3 -1 -4 -3 1 ...
This satisfies the relation and does not have it true for any p < 6. Thus p = 6 and the answer is d.

Answer 2

g(x) = g(x+1) + g(x-1)
for g(x) = g(x+p)
g(x+p) = g(x-1) + g(x+1)
since g(x+1) > g(x-1) with 2 steps
p = 2

Answer 3

(c) 2, as you can see by comparing g(1) to g(3).

Answer 4

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