That simple (if I am not wrong somewhere)? Let $a_n=\frac{p_n}{q_n}$ with $(p_n, q_n)=1$. Hence $q_np_{n+1}(p_{n+1}-q_{n+1})=p_nq_{n+1}^2$, so $q_n | q_{n+1}^2 | q_n$ and thus $q_n=q_{n+1}^2$ so $q_i=1$ for all $i$. So $a_n$ shall be a positive integer. Note that $(2a_{n+1}-1)^2=4a_n+1$, so if $4a_n+1=b_n^2$, then $b_{n+1}^2=2b_n+3$. Note that this sequence is decreasing, hence eventually constant, which shall be $3$. Hence $a_n=2$ for all $n$ and the only value for $a_1$ is $2$.

VicKmath7 wrote: That simple (if I am not wrong somewhere)? Let $a_n=\frac{p_n}{q_n}$ with $(p_n, q_n)=1$. Hence $q_np_{n+1}(p_{n+1}-q_{n+1})=p_nq_{n+1}^2$, so $q_n | q_{n+1}^2 | q_n$ and thus $q_n=q_{n+1}^2$ so $q_i=1$ for all $i$. So $a_n$ shall be a positive integer. Note that $(2a_{n+1}-1)^2=4a_n+1$, so if $4a_n+1=b_n^2$, then $b_{n+1}^2=2b_n+3$. Note that this sequence is decreasing, hence eventually constant, which shall be $3$. Hence $a_n=2$ for all $n$ and the only value for $a_1$ is $2$. i don't understand "Note that $(2a_{n+1}-1)^2=4a_n+1$, so if $4a_n+1=b_n^2$, then $b_{n+1}^2=2b_n+3$. Note that this sequence is decreasing, hence eventually constant, which shall be $3$. Hence $a_n=2$ for all $n$ and the only value for $a_1$ is $2$."

VicKmath7 wrote: i don't understand "Note that $(2a_{n+1}-1)^2=4a_n+1$, so if $4a_n+1=b_n^2$, then $b_{n+1}^2=2b_n+3$. Note that this sequence is decreasing, hence eventually constant, which shall be $3$. Hence $a_n=2$ for all $n$ and the only value for $a_1$ is $2$." Since he managed $4a_n+1 = b_n^2$ then $4a_{n+1}+1 = b_{n+1}^2$ or $4a_{n+1}-2 = b_{n+1}^2-3$ Which implies $2a_{n+1}-1= \frac{b_{n+1}^2-3}{2}$ Then $L.H.S. = \frac{(b_{n+1}^2-3)^2}{4} = R.H.S = b_n^2$ Thus $\frac{b_{n+1}^2-3}{2} = b_n$ And yeah, the sequence is constantly at some point Let $b_k = 3$ We observed that $b_i = 3 $ for all positive integer i If I understand correctly.

Quidditch wrote: Determine all possible values of $a_1$ for which there exists a sequence $a_1, a_2, \dots$ of rational numbers satisfying $$a_{n+1}^2-a_{n+1}=a_n$$for all positive integers $n$. $\color{blue} \boxed{\textbf{Answer:0, 2 are the only solutions}}$ $\color{blue} \boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$a_n=\frac{x_n}{y_n}, x_n, y_n\in\mathbb{Z}, gcd(x_n,y_n)=1$$$$\Rightarrow a_{n+1}^2-a_{n+1}-\frac{x_n}{y_n}=0$$$$\Rightarrow y_na_{n+1}^2-y_na_{n+1}-x_n=0$$By Rational Root Theorem: $$\Rightarrow x_{n+1}|x_n, y_{n+1}|y_n$$$$\Rightarrow \exists M\in\mathbb{Z^+} / |a_i|=|a_j|, \forall i,j\in\mathbb{Z}_{>M}$$$$\Rightarrow a_n^2\pm a_n=a_n, \forall n\in\mathbb{Z}_{>M}$$$$\Rightarrow a_n=0 \text{ or }2, \forall n\in\mathbb{Z}_{>M}$$Now we simply go backwards, let $z \in \mathbb{Z^+}$ be the smallest index for which the sequence becomes constant. $$\Rightarrow a_z^2-a_z=a_{z-1}$$$\color{red} \boxed{\textbf{If it becomes constant 0:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow a_{z-1}=0$$$$\Rightarrow a_i=0, \forall i\in\mathbb{Z}$$$$\Rightarrow a_1=0$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red} \boxed{\textbf{If it becomes constant 2:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow a_{z-1}=2$$$$\Rightarrow a_i=2, \forall i\in\mathbb{Z}$$$$\Rightarrow a_1=2$$$\color{red}\rule{24cm}{0.3pt}$ $\color{blue}\rule{24cm}{0.3pt}$ $\color{blue}\boxed{\textbf{Conclusion:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$\boxed{\textbf{0, 2 are the only solutions}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$