Exercise 1.2.6 (Product of four consecutive integers)

Proof.

(a)

By the Euclidean division, every integer $n$ is of the form $n=3k+r$, where $r=0,1$, or $2$.

If $n=3k$, then $3\mid n$; if $n=3k+1$, then $3\mid n+2$; if $n=3k+2$, then $3\mid n+1$. In every case,

$$3\mid n(n+1)(n+2).$$

Moreover, since $n$ is odd or even, $2\mid n$ or $2\mid n+1$. In both cases, $2\mid n(n+1)$, a fortiori

$$2\mid n(n+1)(n+2).$$

Moreover, $2$ and $3$ are relatively prime, therefore, for every integer $n$,

$$5\mid n(n+1)(n+2).$$

The product of three consecutive integers is divisible by $6$.

(b)

Similarly, every integer $n$ is of the form $n=4k+r$, where $r=0,1,2$, or $4$.

If $r=0$, $4\mid n$ and $2\mid n+2$, so $8\mid n(n+2)$;

if $r=1$, $4\mid n+3$ and $2\mid n+1$, so $8\mid (n+1)(n+3)$;

if $r=2$, $4\mid n+2$ and $2\mid n$, so $8\mid n(n+2)$;

if $r=3$, $4\mid n+1$ and $2\mid n+3$, so $8\mid (n+1)(n+3)$.

In every case

$$8\mid n(n+1)(n+2)(n+3).$$

By part (a), $3\mid n(n+1)(n+2)$, a fortiori

$$3\mid n(n+1)(n+2)(n+3).$$

Since $\gcd (3,8)=1$,

$$24\mid n(n+1)(n+2)(n+3).$$

The product of four consecutive integers is divisible by $24$.