[ $\underline{\text { U }}$ equations in integers ]

Solve the equations in natural numbers:

a) $x^{2}-y^{2}=31$

b) $x^{2}-y^{2}=303$.

#
Factorize the left-hand side.

## Solution

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

a) Since the product is equal to the prime number 31, the larger factor is 31, and the smaller one is 1. Thus, $x-y=1, x+y=31$, from which we get $x=16, y=15$.

b) $303=1 \cdot 303=3 \cdot 101$. We have two cases.

1) $x-y=1, x+y=303$, from which we get $x=152, y=151$.
2) $x-y=3, x+y=101$, from which we get $x=52, y=49$.

## Answer

a) $(16,15)$; b) $(152,151)$, ( 52,49$)$.