223. For what values of $n$ is the number $n^{2}-1$ the product of four consecutive natural numbers?
223. Let $\quad n^{2}-1=x(x+1)(x+2)(x+3), \quad\left(x^{2}+\right.$ $+3 x)^{2}+2\left(x^{2}+3 x\right)+1=n^{2}, \quad x^{2}+3 x+1=n, \quad x=$ $=\frac{-3+\sqrt{5+4 n}}{2} ;$ hence, $5+4 n=z^{2}$; from this it is clear that $z^{2}-5$ and, therefore, $z^{2}-1$ is divisible by 4; and since $z^{2}-1=(z+1)(z-1)$, it is immediately clear that $z$ is an odd number, $z=2 t+1 ;$ therefore, $n=\frac{z^{2}-5}{4}=t(t+1)-$ $-1, x=t-1$.

Answer. The number $n^{2}-1$ is the product of four consecutive natural numbers under the condition $n=$ $=t(t+1)-1 ;$ then $n^{2}-1=(t-1) t(t+1)(t+2)$.