Find all pairs of positive integers $(a, b)$ satisfying the following conditions: (i) $a$ divides $b^{4}+1$, (ii) $b$ divides $a^{4}+1$, (iii) $\lfloor\sqrt{a}\rfloor=\lfloor\sqrt{b}\rfloor$.
Obviously, $\operatorname{gcd}(a, b)=1$, so the problem conditions imply 
$$ a b \mid(a-b)^{4}+1 $$
since each of $a$ and $b$ divide the right-hand side. We define 
$$ k \stackrel{\text { def }}{=} \frac{(b-a)^{4}+1}{a b} . $$
Claim (Size estimate) — We must have $k \leq 16$.
$$ \begin{aligned} a b & \geq n^{2}\left(n^{2}+1\right) \geq n^{4}+1 \\ (b-a)^{4}+1 & \leq(2 n)^{4}+1=16 n^{4}+1 \end{aligned} $$
which shows $k \leq 16$.
Claim (Orders argument) - In fact, $k=1$.
Thus $k$ is odd. However, every odd prime divisor of $(b-a)^{4}+1$ is congruent to 1 $(\bmod 8)$ and is thus at least 17, so $k=1$ or $k \geq 17$. It follows that $k=1$.
At this point, we have reduced to solving 
$$ a b=(b-a)^{4}+1 $$
$$ a^{2}-d a-\left(d^{4}+1\right)=0 $$
The discriminant $d^{2}+4\left(d^{4}+1\right)=4 d^{4}+d^{2}+4$ must be a perfect square.
- The cases $d=0$ and $d=1$ lead to pairs $(1,1)$ and $(1,2)$.
- If $d \geq 2$, then we can sandwich 
$$ \left(2 d^{2}\right)^{2}<4 d^{4}+d^{2}+4<4 d^{4}+4 d^{2}+1=\left(2 d^{2}+1\right)^{2} $$
so the discriminant is not a square.
Remark (Author remarks on origin). This comes from the problem of the existence of a pair of elliptic curves over $\mathbb{F}_{a}, \mathbb{F}_{b}$ respectively, such that the number of points on one is the field size of the other. The bound $n^{2} \leq a, b<(n+1)^{2}$ is the Hasse bound. The divisibility conditions correspond to asserting that the embedding degree of each curve is 8, so that they are pairing friendly. In this way, the problem is essentially the key result of https://arxiv.org/pdf/1803.02067.pdf, shown in Proposition 3.