Diophantine polynomial equations

Question

If K is a positive integer, what is the smallest k for which $x^2 + kx = 4y^2 -4y +1$ has a solution (x,y) where x and y are integers

= $(2y-1)^2 = x^2 +kx$

=$(2y-1+x)(2y-1-x) = kx $

I tried equating them to each other

$(2y-1+x) = k; (2y-1-x) = x$

$x= k-2y+1; k= (6y-3)/2$

but then k does not have a solution.

Answer 1

If $k$ is odd, the equation can not hold since the LHS is even and the RHS is odd. Now let $k' := \frac k2$ and $x' := x + k'$ and $y' := 2y-1$.

After arrangement $x'^2 = y'^2 + k'^2$ with $y'$ being odd. We know that the smallest $k'$ is 4 so $k$ is at least 8.

Answer 2

We need $$k^2+(4y-2)^2$$ to be perfect square

WLOG $k=a(p^2-q^2),2(2y-1)=a(2pq)$

$\implies apq$ is odd

The minimum positive values of $a,p^2,q^2$ will be $1,3^2,1^2$ respectively

$$x=\dfrac{a(q^2-p^2)\pm a(p^2+q^2)}2=?$$