Given, f(x+1)+f(x−1)=√2f(x)...(i) Replace x by x+1 f(x+2)+f(x)=√2f(x+1)...(ii) Replace x by x−1 in Eq. (i), f(x)+f(x−2)=√2f(x−1)...(iii) On adding Eqs. (ii) and (iii), f(x+2)+f(x−2)‌+2f(x) ‌=√2[f(x+1)+f(x−1)] ⇒f(x+2)+f(x−2)‌=√2(√2f(x))−2f(x)...[from‌Eq.(i)] ∴f(x+2)+f(x−2)=0