2. determine the two smallest natural numbers that can be written in the form $7 m^{2}-11 n^{2}$ with natural numbers $m$ and $n$.
## 1st solution

Assume that $7 m^{2}-11 n^{2}=c$ applies with a natural number $c$.

We consider the equation modulo 7, which means that $c \equiv-11 n^{2} \equiv 3 n^{2}(\bmod$ 7). Because of $n^{2} \equiv 0,1,2,4$ is therefore $c \equiv 0,3,5,6(\bmod 7)$. Similarly, the equation $c \equiv 7 m^{2}$ is obtained modulo 11 and because of $m^{2} \equiv 0,1,3,4,5,9$ thus $c \equiv 0,2,6,7,8,10(\bmod 11)$. Modulo 4 finally holds $c \equiv 3\left(m^{2}-n^{2}\right) \equiv 0,1,3$. These congruences exclude many values for $c$. The two smallest remaining possibilities are $c=7.13$.

We now show conversely that $c=7$ and $c=13$ are assumed.

- For every solution $(m, n)$ of the equation $7 m^{2}-11 n^{2}=7$, $n$ must be divisible by 7. Set $n=7 k$, then $(m, k)$ is a solution of

$$
m^{2}-77 k^{2}=1
$$

and conversely, each solution of (1) yields one of the original equation. However, (1) is a Pell equation and is therefore known to have an infinite number of solutions.

- Through trial and error you can easily find

$$
7 \cdot 4^{2}-11 \cdot 3^{2}=13
$$

Note: The minimum solution of the Pell equation $(1)$ is $(m, k)=(351,40)$ and thus the smallest possible solution $7 \cdot 351^{2}-11 \cdot 280^{2}=7$ is obtained for $c=7$.