Find all pairs $(m,n)$ of natural numbers with $m<n$ such that $m^2+1$ is a multiple of $n$ and $n^2+1$ is a multiple of $m$.
1. Given the conditions \( n \mid m^2 + 1 \) and \( m \mid n^2 + 1 \), we start by noting that \( \gcd(m, n) = 1 \). This is because if \( d \) is a common divisor of \( m \) and \( n \), then \( d \) must also divide both \( m^2 + 1 \) and \( n^2 + 1 \). However, since \( d \) divides both \( m \) and \( n \), it must divide their linear combinations, leading to a contradiction unless \( d = 1 \).

2. Since \( \gcd(m, n) = 1 \), we have \( mn \mid m^2 + n^2 + 1 \). Let the quotient of this division be \( k \), i.e., 
   \[
   m^2 + n^2 + 1 = kmn.
   \]
   Rearranging this equation as a quadratic in \( m \), we get:
   \[
   m^2 - (kn)m + n^2 + 1 = 0.
   \]

3. If there exists a solution \((m, n)\), then \( n \) is a root of the quadratic equation. Let \( n_0 \) be the other root. By Vieta's relations, we have:
   \[
   n + n_0 = kn,
   \]
   \[
   nn_0 = m^2 + 1.
   \]

4. From the first equation, we get:
   \[
   n_0 = kn - n.
   \]
   Substituting this into the second equation, we get:
   \[
   n(kn - n) = m^2 + 1,
   \]
   \[
   kn^2 - n^2 = m^2 + 1,
   \]
   \[
   n^2(k - 1) = m^2 + 1.
   \]

5. Since \( m < n \), we have:
   \[
   nn_0 = m^2 + 1 < n^2 \implies n_0 < n.
   \]

6. If \((m, n)\) is a solution, then \((m, n_0) = (m, kn - n)\) is another solution with \( n_0 < n \). Notice that:
   \[
   kn = n + n_0 > 2n_0.
   \]
   This implies \( k > 1 \), so \( m > n_0 \).

7. This means we can repeat the steps with the new pair \((m, n_0)\), but this time swapping \( n \to m \) in the quadratic equation. This process will eventually lead to a minimal solution \((m, n)\).

8. The condition \( nn_0 = m^2 + 1 \) ensures that we can always find a smaller \( n_0 \) if \( n > 1 \). Therefore, the only possible minimal solution is \((m, n) = (1, 1)\).

The final answer is \( \boxed{ (m, n) = (1, 1) } \).