Let f:N→N where N is set of natural numbers be a function such that f(x+y)=f(xy)∀x≥4 and y≥4, then
f(4)=f(5)
f(8)=f(9)
f(7)=f(8)
f(5)=f(6)
f(1+f(1))=4f(1)⇒ f(1+4)=4×4⇒ f(5)=16 Now f(5+f(5))=4f(5)⇒ f(5+16)=4×16⇒f(21)=64