Sum of smallest differences

It is clear that $x>y$ and $x-y$ is a natural number. Let $z=x-y$ , equation becomes $4(y+z)^2+y+z=5y^2+y \\ 4y^2+8yz+4z^2+z=5y^2 \\ y^2-y(8z)-(4z^2+z)=0$ This equation is a quadratic in terms of $y$ and its discriminant must be a perfect square, hence $\Delta'=(4z)^2+(4z^2+z)=20z^2+z=z(20z+1)=n^2$ Since $\gcd(z,20z+1)=1$ and product of them is a perfect square both of them must be perfect squares which means $z=b^2 \\20z+1=a^2$ Combining last two equations gives $20b^2+1=a^2$ or $a^2-20b^2=1$ Which is the well-known Pell's equation. Fundamental solution of the equation is $(a_1,b_1)=(9,2)$ and $a_k = \frac{1}{2}\left((9 + 2\sqrt{20})^k + (9 -2 \sqrt{20})^k \right) \\ b_k = \frac{1}{2 \sqrt{20}}\left((9 + 2\sqrt{20})^k - (9 -2 \sqrt{20})^k \right)$ Which gives us $(a_2,b_2)=(161,36)$ and $(a_3,b_3)=(2889,646)$ . Therefore $N=\sum_{k=1}^{3}(x-y)_k=\sum_{k=1}^{3}z_k=b_1^2+b_2^2+b_3^2=2^2+36^2+646^2=418616$ Sum of digits of $418616$ is $26$ .

$4x^2 + x = 5y^2 + y$ , where $x$ and $y$ are natural numbers.

Solve for x using the quadratic formula: $x = \frac{-1 \pm \sqrt{80y^2 + 16y + 1}}{8}$

$x$ cannot be negative so: $x = \frac{-1 + \sqrt{80y^2 + 16y + 1}}{8}$

If $x$ is an integer, then $\sqrt{80y^2 + 16y + 1} \equiv 1$ (mod $8$ ) and $80y^2 + 16y + 1 = m^2$ , where m is natural number.

Solve for $y$ using the quadratic formula: $y = \frac{-16 \pm \sqrt{320m^2 - 64}}{160}$

$y$ cannot be negative so: $y = \frac{-16 + \sqrt{320m^2 - 64}}{160} = \frac{-2 + \sqrt{5m^2 - 1}}{20}$

If $y$ is an integer then $\sqrt{5m^2 - 1} = 2$ (mod $20$ ) and $5m^2 - 1 = s^2$ , where s is natural number.

$5m^2 - 1 = s^2 \Rightarrow s^2 - 5m^2 = -1$

Now we have a Pell's Equation.

We can easily find that $s = 2, m = 1$ is the fundamental solution.

By using the fundamental solution we can find that the $7$ smallest solutions for m are $(1, 17, 305, 5473, 98209, 1762289, 31622993)$

$\sqrt{5m^2 - 1} = 2$ (mod $20$ ) must be true so the $4$ smallest solutions are $(1, 305, 98209, 31622993)$

Now corresponding values of y are $(0, 34, 10980, 3535558)$ .

$y$ cannot be $0$ so possible values are $(34, 10980, 3535558)$

$\sqrt{80y^2 + 16y + 1} \equiv 1$ (mod $8$ ) is true for the $3$ values of $y$

Now corresponding values of $x$ are (38, 12276, 3952874)

Now $N = (38 - 34) + (12276 - 10980) + (3952874 - 3535558) = 418616$

Sum of the digits of $N = 4 + 1 + 8 + 6 + 1 + 6 = \boxed {26}$