$ f:\mathbb{N}\to\mathbb{N}. \\ \ \\\text{i)} \ \ \ \ f(x-1)f(x+1)=\big(f(x)\big)^2 \text{ for } \forall x \in \mathbb{N} \text{ s.t. } \frac x 2 \in \mathbb{N}. \\ \text{ii)} \; \; f(x-1)+f(x+1)=2f(x) \text{ for } \forall x \in \mathbb{N} \text{ s.t. } \frac x 2 \not\in \mathbb{N}. \\ \text{iii)} \ f(2)=2f(1) \\ \ \\ \text{Find } f \text{ with Mathematical Induction of the value of } f(x)-f(x-1).$
Here is my solution, which is quite long and hard to read. I'm posting my solution, but I'm waiting for another clear solution for this.
\begin{align} \ \\ \ \\ \text{Claim: } & f(x)-f(x-1) = \bigg[ \frac {x+1} 2 \bigg]f(1). \ \text{($[x]$: greatest integer function.)} \\ \ \\ x=2; \ & f(2)-f(1)=2f(1)-f(1)=f(1)=\bigg[ \frac 3 2 \bigg]f(1). \\ \ \\ \text{Assume) } & \text{The Claim is valid until } x=k. \\ \ \\ x=k+1; \ & \Bigg( \text{WTS) } f(k+1)-f(k)=\bigg[ \frac {k+2} 2 \bigg]f(1). \Bigg) \\ \text{(i)} \ \frac k 2 \not\in \mathbb{N}: \ & \bigg[ \frac {k+1} 2 \bigg] = \bigg[ \frac {k+2} 2 \bigg]. \\ & f(k-1)+f(k+1)=2f(k). \\ \therefore \ & f(k+1) =(f(k)-f(k-1))+f(k) \\ & = \bigg[ \frac {k+1} 2 \bigg]f(1)+f(k) \\ & = \bigg[ \frac {k+2} 2 \bigg]f(1)+f(k). \\ \ \\ (ii) \ \frac k 2 \in \mathbb{N}: \ &\bigg[ \frac {k+1} 2 \bigg] = \bigg[ \frac {k+2}{2} \bigg]-1. \\ & f(k-1)f(k+1)=f(k)^2. \\ \therefore \ & f(k+1)\Bigg( f(k)-\bigg[ \frac {k+1} 2 \bigg]f(1)\Bigg)=f(k)^2. \\ \Rightarrow\ & f(k)\Big(f(k+1)-f(k)\Big)=\bigg[ \frac {k+1} 2 \bigg] f(1)f(k+1). \\ \Rightarrow \ & f(k+1)-f(k)=\bigg[ \frac {k+1} 2 \bigg] f(1) \frac {f(k+1)}{f(k)} \\ & = \bigg[ \frac {k+1} 2 \bigg] f(1)\Bigg( 1 + \frac {\big[ \frac {k+1} 2 \big]f(1)} {f(k-1)} \Bigg) \\ & = \Bigg( \bigg[ \frac {k+1} {2} \bigg]+\frac {\big[ \frac {k+1} 2 \big]^2f(1)} {f(k-1)} \Bigg) f(1) \\ & = \Bigg( \bigg[ \frac {k+2} 2 \bigg]-1+\frac {\Big( \big[ \frac {k+2} 2 \big]^2-2\big[ \frac {k+2} 2 \big]+1 \Big) f(1)} {f(k-1)} \Bigg) f(1) \\ & = \Bigg( \bigg[ \frac {k+2} 2 \bigg] + \frac {\frac {k^2-4k+4} 4 -k-2+1} {f(k-1)} - 1 \Bigg)f(1) \\ & = \Bigg( \Bigg[ \frac {k+2} 2 \Bigg]+\frac {k^2f(1)} {4f(k-1)}-1 \Bigg) f(1). \\ \ \\ &\begin{matrix} & f(k-1)-\color{Red}{f(k-2)} & = & \Bigg[ \dfrac k 2 \Bigg]f(1) \\ & \color{Red}{f(k-2)}-\color{Blue}{f(k-3)} & = & \bigg[ \dfrac {k-1} 2 \bigg]f(1) \\ & \color{Blue}{f(k-3)}-\color{Red}{f(k-4)} & = & \bigg[ \dfrac {k-2} 2 \bigg]f(1) \\ & \vdots & \vdots & \vdots \\ {\huge +} & \color{Red}{f(2)}-f(1) & = & \bigg[ \dfrac 3 2 \bigg]f(1). \\ {\Huge-} & {\Huge ----} & {\Huge --} & {\Huge ---} \\ & f(k-1)-f(1) & = &\Bigg(\displaystyle \sum_{i=3}^k \bigg[ \dfrac k 2 \bigg] \Bigg) f(1) \end{matrix} \\ \therefore \ & f(k-1)=f(1)+\Bigg( \displaystyle \sum_{i=2}^{\frac k 2} \bigg[ \frac {2i-1} 2 \bigg] + \bigg[ \frac {2i} 2 \bigg] \Bigg) f(1) \\ & = f(1)+\Bigg( \displaystyle \sum_{i=2}^{\frac k 2} (i-1)+i \Bigg) f(1) \\ & = \Bigg( \displaystyle \sum_{i=1}^{\frac k 2} 2i-1 \Bigg)f(1) \\ & = \Bigg( \bigg( \frac k 2 \bigg) \bigg( \frac k 2 + 1 \bigg)-\frac k 2 \Bigg)f(1) \\ & = \frac {k^2f(1)} 4. \\ \therefore \ & 4f(k-1)=k^2f(1). \label{(1)} \tag{1}\\ \ \\ \therefore \ & f(k+1)-f(k)=\Bigg( \bigg[ \frac {k+2} 2 \bigg] + \frac {k^2f(1)} {4f(k-1)} - 1 \Bigg)f(1) \\ & = \bigg[ \frac {k+2} 2 \bigg]f(1). \\ \ \\ \therefore \ & \text{Claim proved.} \\ \ \\ f(x) & = f(x-1)+\bigg[ \frac {x+1} 2 \bigg]f(1) \\ & = f(x-2)+ \Bigg( \bigg[ \frac x 2 \bigg] + \bigg[ \frac {x+1} 2 \bigg] \Bigg) f(1) \\ & \ \ \vdots \\ & = f(1) + \Bigg(\displaystyle \sum_{k=3}^{x+1} \bigg[ \frac k 2 \bigg] \Bigg) f(1) \\ & = f(1)\sum_{k=1}^{x+1} \bigg[ \frac k 2 \bigg]. \Bigg( \because \bigg[ \frac 1 2 \bigg] = 0, \bigg[ \frac 2 2 \bigg] = 1. \Bigg) \\ \therefore \ & f(x)=c\sum_{k=1}^{x+1} \bigg[ \frac k 2 \bigg]. \\ \text{By $\color{#4995CF}{(\ref{(1)})}$, }& f(x)=\frac {(x+1)^2} 4 f(1) \text{ if $x$ is odd.} \\ \text{if $x$ is even: } & f(x)=f(x-1)+\bigg[ \frac {x+1} 2 \bigg]f(1) \\ & = \frac {x^2} {4} f(1) + \frac x 2 f(1) \\ &=\bigg( \frac x 2 \bigg) \bigg( \frac x 2 +1 \bigg) f(1). \\ \ \\ \therefore \ & f(x)= \begin{cases} \bigg( \dfrac {x+1} {2} \bigg)^2c & \text{if $x$ is odd.} \\ \bigg(\dfrac x 2 \bigg) \bigg( \dfrac x 2 +1 \bigg)c & \text{if $x$ is even.} \end{cases} \end{align}