Prove that $f(x)=8$ for all natural numbers $x\ge{8}$ A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
 A: For $x\geq 4$, we have $$f(x+5)=f(5x)=f(4x+x)=f(4x^2)=f(2x\cdot 2x)=f(4x)=f(x+4)$$ so $f$ is constant over $[8,\infty)$ as desired.
A: Yes. It is true that $f(x) = 8 \quad \forall\;\;\; x \in N $
Manually, we can prove this for $x \le 20$.
Now, let $x$ be even. $x = 2y$ for some $y$.
$$f(2y)=f((2y-4) +(4))=f(4(2y-4))=f(8(y-2))=f(8+y-2)=f(y+6)$$
Note: This is true only if the $y-2$ factor is greater than $4$, so let $y \ge 6$.
Similarly, if $x$ is odd, $x = 2y + 1$ for some $y$.
$$f(2y+1)=f((2y-4)+5)=f(5(2y-4))=f(10(y-2))=f(10 + y-2)=f(y+8)$$
Note: Similarly, this has the same condition $y \ge 6$.
And we can see that $2y > y+6$ and $2y+1 > y+8$ for $y\ge6$. ($y > 7$ for the second case). Therefore, for any $f(m)$, we can find $f(n)=f(m)$ for $n < m
$.
Thus after reducing, we get a number lesser than 20 which can be proved manually equal to $8$.
Therefore $f(x) = 8\;\;\; \forall \;\;\; x \in N$
A: $\>$$\>$$\>$$\>$$\>$SS_C4's answer has two problems; the untrue claim that $2y>y+6$, for $y\geq 6$, and not demonstrating that $f(x)=f(8)$, for $9\leq x<17$, but SS_C4's proof can be saved as follows. The basic ideas belong to SS_C4. The value of $f(8)$ is immaterial. Lower-case Latin letters, except $f$, denote positive integers. 
(A) $f(9)=f(4+5)=f(20)=f(4+16)=f(64)=f((8)(8))=f(16)=f((4)(4))=f(8)$.
(B) $f(10) = f(5+5)=f(25)=f(5+20)=f(100)=f((10)(10))=f(20)=f(8)$, by (A).
(C) $f(11)=f(4+7)=f(28)=f(4+24)=f(96)=f((8)(12))=f(20)=f(8)$, by (A).
(D) $f(12)=f(5+7)=f(35)=f(5+30)=f(150)=f((10)(15))=f(25)=f(8)$, by (B).
(E) $f(13)=f(5+8)=f(40)=f((4)(10))=f(14)=f(6+8)= f(48)=f((4)(12))=$
   $\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$f(16)=f(8)$, by (A).
(F) $f(15)=f(4+11)=f(44)=f(4+40)=f(160)=f((8)(20))=f(28)=f(8)$, by (C).
$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$Suppose $f(x) \neq f(8)$, for some $x\geq9$. Let $m=$Min$\{x|x\geq9$ and $f(x)\neq f(8)\}$.
$m \geq 17$, by (A) - (F).
$\>$$\>$$\>$$\>$$\>$$\>$Suppose $m$ is even. Then, $m=2y$. Thus, $y\geq 9, y-2\geq7, 2y-4\geq14$ and
$15\leq y+6<2y=m$. Therefore, by the definition of $m$,
$\>$$\>$$\>$$\>$$\>$$\>$$\>$$f(8)=f(y+6)=f(8+(y-2))=f(8(y-2))=f(4(2y-4))=f(2y)=f(m)$,
contradicting the definition of $m$. Thus, $m$ is odd. So, $m=2y+1$. Therefore,
 $y\geq 8,y-2\geq6, 2y-4\geq12$ and $16\leq y+8<2y+1=m$. Hence, by the definition of $m$,
$\>$$\>$$\>$$\>$$\>$$\>$$f(8)=f(y+8)=f(10+(y-2))=f(10(y-2))=f(5(2y-4))=$
$\>$$\>$$\>$$\>$$\>$$\>$$f(5+(2y-4))=f(2y+1)=f(m),$
again contradicting the definition of $m$. Thus, $f(x)=f(8)$, for $x\geq9$. 
