<s>[INST] 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 $x\ge{8}$
 [/INST] 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 $f(x) = 8\;\;\; \forall \;\;\; x \in N$
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