8.3. What is the smallest natural number $n$ for which there exist natural numbers $x$ and $y$ satisfying the equation a) $x \cdot(x+n)=y^{2}$; b) $x \cdot(x+n)=y^{3}$?
Answer: a) $n=3$; b) $n=2$. Solution. a) For $n=3$ such $x$ and $y$ exist: $x=1, y=2$. The impossibility of the values $n=1$ and $n=2$ follows from the inequalities $x^{2}<x(x+1)<(x+1)^{2}$ and $x^{2}<x(x+2)<(x+1)^{2}$. b) For $n=2$ such $x$ and $y$ exist: $x=2, y=2$. For $n=1$, from the equation $x(x+1)=y^{3}$ it follows that two consecutive natural numbers $x$ and ($x+1$) are cubes of natural numbers, since $x$ and ($x+1$) are coprime numbers (indeed: if the product $ab$ of two coprime numbers $a$ and $b$ is a perfect cube, then the cubes of prime numbers, entering the factorization of this product into prime factors, enter the factorization of only one of these numbers - $a$ or $b$). However, it is obvious that this is impossible (the cubes of consecutive natural numbers differ by at least $7=2^{3}-1^{3}$).