[ Factorization ]  [ Prime numbers and their properties ]

Knowing that the number 1993 is prime, determine whether there exist natural numbers $x$ and $y$ such that
a) $x^{2}-y^{2}=1993$

b) $x^{3}-y^{3}=1993$

c) $x^{4}-y^{4}=1993$?
Factorize the left side.

## Solution

a) $(x-y)(x+y)=1993$. Considering that the number 1993 is prime, we get $x-y=1, x+y=1993$. Hence, $x=997, y=996$.

b) $(x-y)\left(x^{2}+x y+y^{2}\right)=1993$, from which $x-y=1, x^{2}+x y+y^{2}=(x-y)^{2}+3 x y=1993$. Therefore, $x y=664=23.83$. There are no solutions.

c) $\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)=1993$, from which $x^{2}-y^{2}=1, x^{2}+y^{2}=1993$. From a) it follows that $x^{2}=997, y^{2}=996$. There are no solutions.

## Answer

a) Exist; b)-c) do not exist.