482. Three peasants, John, Peter, and Alexis, came to the market with their wives, Mary, Kitty, and Jenny. To find out who is married to whom, if it is known that each of these six individuals paid as many pence for each item bought as the number of items they bought. Each man spent 63 pence more than his wife. In addition, John bought 23 more items than Kitty, and Peter bought 11 more items than Mary.

## Catalan Problems.
482. Let each man buy $x$ items, so he paid $x^{2}$ pence for them. His wife bought $y$ items and paid $y^{2}$ pence. According to the condition

$$
x^{2}-y^{2}=(x+y)(x-y)=63
$$

Since $x$ and $y$ are integers, the following combinations are possible:

$$
63 \cdot 1=21 \cdot 3=9 \cdot 7
$$

We have:

1) $x+y=63$
2) $x+y=21$;
3) $x+y=9$;
$x-y=1$
$x-y=3$
$x-y=7$
4) $x=32 ; \quad y=31$;
5) $x=12 ; \quad y=9$;
6) $x=8 ; \quad y=1$. 206

The difference of 23 can only be between 32 and 9, so John bought 32 items, and Kitty bought 9. The difference of 11 can only be between 12 and 1, so Peter bought 12 items, and Mary bought one. Finally, Alexis bought 8 items, and Jenny bought 31. It is clear that John's wife was Jenny, Peter's wife was Kitty, and Alexis's wife was Mary.

Catalan, Eugène (1814-1894), a Belgian mathematician who, despite his work in higher mathematics, devoted much attention to elementary mathematics. His "Mélanges mathématiques" (3 volumes) are particularly valuable. The problems presented here are taken from "Théorèmes et problèmes de géométrie élémentaire".