Prove that $4x^{n} + (x+1)^{2} = y^2$ doesn't have positive integer solutions for $n \ge 3$ Equation is: $4x^{n} + (x+1)^{2} = y^2$
Task is to prove:
a) for $n = 1$ there are no solutions
b) for $n = 2$ there is infinite number of solutions
c) for $n \ge 3$ there are no solutions
I have proved $a$ and $b$, but have no idea about $c$. There is nothing common with $a$ and $b$
 A: For $x, y$ positive and $n \geq 3$, note $y \equiv x+1 \pmod{2}$ so 
$$4x^n=y^2-(x+1)^2=(y-(x+1))(y+(x+1))$$
$$x^n=\frac{y-(x+1)}{2}\frac{y+(x+1)}{2}$$
Now if $\exists$ prime $p$ with $p \mid \frac{y-(x+1)}{2}$ and $p \mid \frac{y+(x+1)}{2}$, then $p \mid x^n$ so $p \mid x$. Also $p \mid \frac{y+(x+1)}{2}-\frac{y-(x+1)}{2}=(x+1)$, a contradiction.
Thus $\gcd(\frac{y-(x+1)}{2}, \frac{y+(x+1)}{2})=1$, so both $\frac{y-(x+1)}{2}$ and $\frac{y+(x+1)}{2}$ are (positive) $n$th powers. We may thus write $$\frac{y-(x+1)}{2}=\alpha^n, \frac{y+(x+1)}{2}=\beta^n, x=\alpha\beta$$ for some positive integers $\alpha, \beta$.
Then $\alpha\beta+1=x+1=\frac{y+(x+1)}{2}-\frac{y+(x+1)}{2}=\beta^n-\alpha^n$.
Clearly we require $\beta>\alpha$. Write $\beta=\alpha+k$, $k$ a positive integer. Then
$$\alpha^2+k\alpha+1=\alpha\beta+1=\beta^n-\alpha^n=(\alpha+k)^n-\alpha^n \geq nk\alpha^{n-1} \geq 3k\alpha^2$$
However 
$$3k\alpha^2-(\alpha^2+k\alpha+1)=(k-1)\alpha^2+k\alpha(\alpha-1)+(k\alpha^2-1) \geq 0$$
so equality holds and we must have 
$$(k-1)\alpha^2=k\alpha(\alpha-1)=(k\alpha^2-1)=0$$
i.e. $k=\alpha=1$, i.e. $\alpha=1, \beta=2$. But then $3=\alpha\beta+1=\beta^n-\alpha^n=2^n-1$, so $n=2$, a contradiction.
