[ Factorization ] [ Equations in integers ]

In how many ways can the number 1979 be represented as the difference of squares of natural numbers?
Let $1979=x^{2}-y^{2}=(x-y)(x+y)$. Since $1979-$ is a prime number, there is a unique representation of it as a product of two natural numbers $1979=1 \cdot 1979$. Since $x-y<x+y$, then $x-y=$ $1, x+y=1979$, from which $x=990, y=989$.

## Answer

One way.