> [!NOTE] Lemma > There does not exist a strictly increasing [[Totally Multiplicative Functions|totally multiplicative function]] $f:\mathbb{N}^+\to \mathbb{N}^+$ such that $f(2)=3.$ ###### Proof Let $k:=f(3).$ Since $2^3<3^2,$ $f(8)=27<k^2$ which gives $k>5.$ Since $3^3<2^5$ so $k^3<3^5=243<7^3$ that is, $k>7.$ Therefore $f(3)=6.$ The monotonicity of $f$ gives that for $u,v\in \mathbb{N}^+,$ $2^u>3^v$ iff $3^u>6^v.$ Taking logs, this gives $\frac{u}{v}>\log_{2} 3$ iff $\frac{u}{v}>\log_{3} 6.$ Since the rationals are dense, we have $\log_{2}3=\log_{3}6.$ So $\log_{2}3 = \frac{\log_{2} 6}{\log_{2}3}=\frac{\log_{2}2+\log_{2}3}{\log_{2}3}=\frac{1}{\log_{2}3}+1.$ That is, $\log_{2}3$ is the positive solution to the quadratic equation $x^2 -x-1=0,$ which is $\frac{1+\sqrt{ 5 }}{2}.$ That is, $2^{\frac{(1+\sqrt{ 5 })}{2}} = 3 $which is equivalent to $2^{1+\sqrt{ 5 }} = 9$But this would imply that $65536=2^{5(3.2)} <2^{5(1+\sqrt{ 5 })}=9^5=59049$ ###### Proof Let $k:=f(3).$ Since $2^3<3^2,$ $f(8)=27<k^2$ which gives $k>5.$ Since $3^3<2^5$ so $k^3<3^5=243<7^3$ that is, $k>7.$ Therefore $f(3)=6.$ However, $3^{8}=6561<8192=2^{13}.$ Thus $6^{8} = f(3^{8})<f(2^{13})=3^{13}$which simplifies to $2^{8}<3^{5}$ but $2^{8}=256$ and $3^{5}= 243.$ # Applications ###### Generalisations For what values of $f(2)$ does a strictly increasing totally multiplicative functions exist? e.g. It exists for $f(2)=4$ - $f(n)=n^{2}.$ Conjecture: $f(n)=n^{k}$ are the only completely multiplicative functions. # References Functional equations: a problem-solving approach (problem 2.3) Putnam & Beyond Q7