Let $2y_n+\frac{3}{2}=a_n$ We have \[a_{n+1}=4a_n-a_{n-1}-1\]where $a_0=1,a_1=\frac{3}{2}$. Denote $x_n=a_n-\frac{1}{2}$ \[x_{n+1}=4x_n-x_{n-1}\]where $x_0=\frac{1}{2}$ and $x_1=1$. Let's show that $x_n=\frac{1}{4}((2+\sqrt3)^n+(2-\sqrt3)^n)$ by induction. It's true for $n=0,1$. Let it be true for $n\leq k$. \[\frac{1}{4}((2+\sqrt3)^{k+1}+(2-\sqrt3)^{k+1})\overset{?}{=}x_{k+1}=4x_k-x_{k-1}=(2+\sqrt3)^k+(2-\sqrt3)^k-\frac{1}{4}((2+\sqrt3)^{k-1}+(2-\sqrt3)^{k-1})\]Let $a=2+\sqrt3$ and $b=2-\sqrt3$ Note that $ab=1$. \[a^{k+1}+b^{k+1}\overset{?}{=}4a^k+4b^k-a^{k-1}-b^{k-1}\iff a^{2k+2}+1\overset{?}{=}4a^{2k+1}+4a-a^{2k}-a^2\]\[(a^{2k}+1)(a^2-4a+1)\overset{?}{=}0\]Which is true since $a^2-4a+1=0$. Thus, $a_n=x_n+\frac{1}{2}=\frac{1}{4}((2+\sqrt3)^n+(2-\sqrt3)^{n}+2)$ \[a_{2n}=\frac{1}{4}((2+\sqrt3)^{2n}+(2-\sqrt3)^{2n}+2)=(\frac{(2+\sqrt3)^n+(2-\sqrt3)^n}{2})^2\]is a perfect square since $(2+\sqrt3)^n+(2-\sqrt3)^n$ is even which can be seen by binomial representation.$\blacksquare$

For a) By, looking at some terms in sequence we deduce that $y_{2k-1}$ is integer (base case $y_1,y_3,y_5$) and $y_{2k-2}$ is form in $\frac{odd}{4}$. Assume till $y_{2k}$ and let's prove for $y_{2k+2}$ and $y_{2k+1}$. So, it is easy to found $y_{2k+1}+y_{2k-1}$ is even thus, we proved $y_{2k+1}$ is even integer. $y_{2k+1}+y_{2k+3}=4y_{2k+2}+1$ that means $y_{2k+2}$ is $\frac{odd}{4}$. Which concludes $2y_{2k}+\frac{3}{2}$ is positive integer (it is positive integer because $y_{n}$ is strictly increasing).

Z4ADies wrote: By, looking at some terms in sequence we deduce that $y_{2k-1}$ is integer (base case $y_1,y_3,y_5$) and $y_{2k-2}$ is form in $\frac{odd}{4}$. Assume till $y_{2k}$ and let's prove for $y_{2k+2}$ and $y_{2k+1}$. So, it is easy to found $y_{2k+1}+y_{2k-1}$ is even thus, we proved $y_{2k+1}$ is even integer. $y_{2k+1}+y_{2k+3}=4y_{2k+2}+1$ that means $y_{2k+2}$ is $\frac{odd}{4}$. Which concludes $2y_{2k}+\frac{3}{2}$ is positive integer (it is positive integer because $y_{n}$ is strictly increasing). We have $y_1=0,y_3=6,y_5=90,y_7=1260$ it is easy to observe $y_{2k+1}=14 \cdot y_{2k-1}-y_{2k-3}$ (can be proven by induction). Then use $2y_{2k}+\frac{3}{2}=\frac{y_{2k-1}+y_{2k+1}+2}{2}$ this formula. Which is finishing problem. I don't see how are you planning to prove that $2y_{2k}+\frac{3}{2}$ is a perfect square