If f(x-1)+f(x+1)=√3f(x) and f(5)=100 for all x€R then find the period of f(x)

Replace x = (x + 2) in above equation,
=> ƒ(x + 2 + 2) + ƒ(x + 2 − 2) = (√2) * ƒ(x + 2)
=> ƒ(x + 4) + ƒ(x) = (√2) * ƒ(x + 2) - - - (1)

Again replace x = (x − 2) in above equation,
=> ƒ(x − 2 + 2) + ƒ(x − 2 − 2) = (√2) * ƒ(x − 2)
=> ƒ(x) + ƒ(x − 4) = (√2) * ƒ(x − 2) - - - (2)

Adding (1) & (2) we get,
=> ƒ(x + 4) + ƒ(x − 4) + 2ƒ(x) = (√2) * ƒ(x + 2) + (√2) * ƒ(x − 2)
=> ƒ(x + 4) + ƒ(x − 4) + 2ƒ(x) = (√2) * [ƒ(x + 2) + ƒ(x − 2)]
=> ƒ(x + 4) + ƒ(x − 4) + 2ƒ(x) = (√2) * (√2) ƒ(x )
. . . . . . . .{As, ƒ(x + 2) + ƒ(x − 2) = (√2) * ƒ(x) }

=> ƒ(x + 4) + ƒ(x − 4) + 2ƒ(x) = 2 ƒ(x )
=> ƒ(x + 4) + ƒ(x − 4) = 0

Now replace, x = (x + 4),
=> ƒ(x + 8) + ƒ(x) = 0
=> ƒ(x + 8) = −ƒ(x) - - - - - - (3)

Replacing x = (x + 8),
=> ƒ(x + 16) = −ƒ(x + 8)
=> ƒ(x + 16) = − [−ƒ(x)]
=> ƒ(x + 16) = ƒ(x)