Let $a=b+k$ then we have: $k-1|2b^2+2bk+k^2$ (1) $19k^2=2b^2+2bk-20$ (2) Rewrite (2) as :$20k^2+20=2b^2+2bk+k^2$ give as $k-1|20(k^2+1)$ which mean $k-1|40$ Now rewrite (2) as: $2b^2+2bk-19k^2-20=0$ $D=4k^2-4*2(-19k^2-20)=A^2$ which gives: $39k^2+40=B^2$ (3) Taking $mod4$ in (3) we have $k=even$ so $k-1=odd$ so $k-1= +-1,+-5$ Now for $k=0,-4$ $39k^2+40$ is not a perfect square We have prove that $k=2or6$ and now by $D$ we can find $b$ and then $a$

it is clear that $ a> b $, then let $ a-b = c $ replacing in the second equation: $$\frac{(b + c) ^ 2 + b ^ 2}{2 (b + c) b-1} = \frac{20}{19}$$, then $ 38b ^ 2 + 38cb + 19c ^ 2 = 40bc + 40c ^ 2-20 $ $\Rightarrow 19c ^ 2 = 2bc + 2b ^ 2-20 $ $\Rightarrow 19c ^ 2 + 20 = 2bc + 2b ^ 2 $ replacing this in the first equation: $ c-1 | c ^ 2 + 19c ^ 2 + 20 $, we know that $ c-1 | c ^ 2-1$, also , $$(c ^ 2-1, c ^ 2 + 1)$$, is at most 2 . $\Rightarrow c-1 | 40$ then we have that c can only be 2 or 6, the rest of the values do not comply when replacing them in the 2 equations. then we have 2 cases: case 1: $a = b + 2$ $\Rightarrow \frac{(b + 2) ^ 2 + b ^ 2}{2 (b + 2) b-1} = \frac{20}{19} $ $\Rightarrow 76 + 76b + 38b ^ 2 = 80b + 40b ^ 2-20 $ $\Rightarrow 48 = b (b + 2) $ $\Rightarrow (a, b) = (8, 6); (-6, -8)$ case 2: $a = b + 6$ $\Rightarrow \frac{(b + 6) ^ 2 + b ^ 2}{2 (b + 6) b-1} = \frac{20}{19} $ $ \Rightarrow (a, b) = (22, 16); (-16, -22) $ Finally the solutions are: $$ (a, b) = (22, 16); (-16, -22); (8, 6); (-6, -8) $$

Let $(1)=a-b-1\mid a^2+b^2$ and $(2)=\frac{a^2+b^2}{2ab-1}=\frac{20}{19}$. From $(2)$ we get $19(a-b)^2=2ab-20$. Let $a=b+k$. We get $19k^2=2b(b+k)-20 (\star)$. From $(1)$ we get $k-1\mid 2b^2+2bk+k^2=20k^2+20 ($beacuse of $(\star))$. Also $(\star)$ gives that $k$ is even. So $k-1\mid 5k^2+5 \implies k-1\mid 10 \implies k-1\mid 5 \implies k=-4,0,2,6$. So checking this values in $(\star)$ gives $(a,b)=(8,6),(-6,-8),(22,16),(-16,-22)$ and indeed these ones works, so we are done.