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) $\lfloor\sqrt{a}\rfloor=\lfloor\sqrt{b}\rfloor$.
The only solutions are $(1,1),(1,2)$, and $(2,1)$, which clearly work. Now we show there are no others.
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 $k \leq 16$.
Proof. Let $n=\lfloor\sqrt{a}\rfloor=\lfloor\sqrt{b}\rfloor$, so that $a, b \in\left[n^{2}, n^{2}+2 n\right]$. We have that
$$
\begin{aligned}
a b & \geq n^{2}\left(n^{2}+1\right) \geq n^{4}+1 \\
(b-a)^{4}+1 & \leq(2 n)^{4}+1=16 n^{4}+1
\end{aligned}
$$
which shows $k \leq 16$.
Claim (Orders argument) - In fact, $k=1$.

Proof. First of all, note that $k$ cannot be even: if it was, then $a, b$ have opposite parity, but then $4 \mid(b-a)^{4}+1$, contradiction.

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
$$
a b=(b-a)^{4}+1
$$
and we need to prove the claimed solutions are the only ones. Write $b=a+d$, and assume WLOG that $d \geq 0$ : then we have $a(a+d)=d^{4}+1$, or
$$
a^{2}-d a-\left(d^{4}+1\right)=0 .
$$

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.
The solution is complete.
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.