a function f(x) satisfies the equation f(x)=f(x-1)+f(x+1) for all values of x. Define f(1)=1 and f(3)=3; then f(2)=1+3=4. find f(1867)?

It is given that: f(x) = f(x - 1) + f(x + 1)

Add two more consecutive relations:
f(x + 1) = f(x) + f(x + 2) …… [1]
f(x + 2) = f(x + 1) + f(x + 3) …… [2]

[1] + [2]:  f(x) + f(x + 3) = 0 …… [3]
Also:  f(x + 3) + f(x + 6) = 0 …… [4]

[3] - [4]:
f(x) - f(x + 6) = 0
f(x + 6) = f(x)

Similarly, f(x + 12) = f(x + 6)
f(x + 18) = f(x + 12)
…… and so on

It can be deduced that f(x + 6n) = f(x)
where n is an integer.

f(1867)
= f(1 + 6×311)
= f(1)
= 1