Question

Show that 1n^3+2n+3n^2 is divided by 2 and 3 for all positive integers n

Answer 1

Answer.  Define $f(n) = 1n^3+2n+3n^2$. We try to factorize this polynomial. One obvious factor is $n$, so $f(n) = n(1n^2+2+3n)$. We factorize the factor of degree $2$ by completing the square:

$$1n^2+2+3n = \left(n+\frac{3}{2}\right)^2 - \frac{9}{4} +2 = \left(n+\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2$$

$$= \left(n +\frac{3}{2} -\frac{1}{2}\right) \left(n +\frac{3}{2} +\frac{1}{2}\right) = (n+1)(n+2).$$

Hence

$$f(n) = n (n+1) (n+2).$$

Let $n$ be a positive integer number. Division of $n$ by $3$ gives an integer quotient $q$ and a remainder $r\in \{0, 1, 2\}$ such that $n = 3q+r$. Now check all the possible values of $r$.

• If $r=0$, then $3$ divides $n = 3q$ which is a factor of $f(n)$.
• If $r=1$, then $3$ divides $n+2 = (3q+1) +2 = 3(q+1)$ which is a factor of $f(n)$.
• If $r=2$, then $3$ divides $n+1 = (3q+2) +1 = 3(q+1)$ which is a factor of $f(n)$.

Hence $3$ divides $f(n)$ in all cases.

Division of $n$ by $2$ gives an integer quotient $q$ and a remainder $r\in \{0, 1\}$ such that $n = 2q+r$.

• If $r=0$, then $2$ divides $n = 2q$ which is a factor of $f(n)$.
• If $r=1$, then $2$ divides $n+1 = (2q+1) +1 = 2(q+1)$ which is a factor of $f(n)$.

Hence $2$ divides $f(n)$ in all cases.