Divisibility test without using PMI

Question

How can we logically show that $n(n+1)(2n+1)$ is divisible by $6$ (without PMI) ?

Answer 1

$$n(n+1)(2n+1)=n(n+1)\{n+2+(n-1)\}$$

$$=\underbrace{(n+2)(n+1)n}_{\text{ Product of three consecutive integers}}+\underbrace{(n+1)n(n-1)}_{\text{ Product of three consecutive integers}}$$

Now use The product of n consecutive integers is divisible by n factorial OR The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

Alternatively, as $2|n(n+1),$ it is sufficient to establish $3|n(n+1)(2n+1)$

Now $n(n+1)(2n+1)=\dfrac{2n(2n+1)(2n+2)}4$

Again the numerator being the product of three consecutive integers, is divisible by $3$

As $(3,4)=1,3$ must divide $\dfrac{2n(2n+1)(2n+2)}4$

Answer 2

An alternative approach:

One of $n$ or $n + 1$ must be even, so $2\mid n(n+1)(2n + 1)$.

Suppose $3\mid n$. Then the result holds, as $2$ and $3$ both divide $n(n+1)(2n + 1)$.

If $n\equiv -1\pmod{3}$, $3\mid n+1$, so $6\mid n(n+1)(2n + 1)$.

The last case is $n\equiv 1\pmod{3}$, in which case $2n+1\equiv 2 + 1\equiv 0\pmod{3}$, so $3\mid 2n+1$, and hence $6\mid n(n+1)(2n + 1)$.

Answer 3

Another alternative:

$$ n(n+1)(2n+1)=2n^3+3n^2+n = 3(n^3+n^2) - (n^3-n)=3(n^3+n^2) - n(n-1)(n+1) $$

The latter is divisibile by $3$ since the first term is obviously a multiple of $3\,$, and the second is the product of $3$ consecutive numbers.

Answer 4

Consider the following cases:

• $n\equiv\color\red0\pmod6 \implies n(n+1)(2n+1)\equiv\color\red0\cdot(\color\red0+1)\cdot(2\cdot\color\red0+1)\equiv0\pmod6$
• $n\equiv\color\red1\pmod6 \implies n(n+1)(2n+1)\equiv\color\red1\cdot(\color\red1+1)\cdot(2\cdot\color\red1+1)\equiv0\pmod6$
• $n\equiv\color\red2\pmod6 \implies n(n+1)(2n+1)\equiv\color\red2\cdot(\color\red2+1)\cdot(2\cdot\color\red2+1)\equiv0\pmod6$
• $n\equiv\color\red3\pmod6 \implies n(n+1)(2n+1)\equiv\color\red3\cdot(\color\red3+1)\cdot(2\cdot\color\red3+1)\equiv0\pmod6$
• $n\equiv\color\red4\pmod6 \implies n(n+1)(2n+1)\equiv\color\red4\cdot(\color\red4+1)\cdot(2\cdot\color\red4+1)\equiv0\pmod6$
• $n\equiv\color\red5\pmod6 \implies n(n+1)(2n+1)\equiv\color\red5\cdot(\color\red5+1)\cdot(2\cdot\color\red5+1)\equiv0\pmod6$