# divisibility application in solving problems

How can I show that $a(2a^2+7)$ is divisible by $3$ for every integer a? I just simplified it as follows.

$$a(2a^2+7) = 3a^3 - a^3 + 9a - 2a = 3a(a^2-3)-a(a^2+2)$$ But I cannot show that $a(a^2+2)$ is a multiple of $3$.

• If $a$ is divisible by $3$, the result is obvious. Show that if $a$ is not divisible by $3$, then $a^2+2$ is divisible by $3$. (Your simplification is nice, but we could also show directly that if $a$ is not divisible by $3$, then $2a^2+7$ is divisible by $3$.) – André Nicolas Apr 2 '14 at 17:34
• here a is an integer. so will it be necessary to for any a ,a^2+2 is divisible by 3 – srimali Apr 2 '14 at 17:36
• ok !. i got it .thanks Bill, a(2a^2+7)=2a^3-2a+9a=2a(a-1)(a+1)+9a .so its obvious now – srimali Apr 2 '14 at 17:39

Hint $\ \color{#c00}{a^3\equiv a}\pmod 3\$ since $\,a^3-a = (a-1)a(a+1)\,$ is the product of $\,3\,$ consecutive integers, one of which must be divisible by $3.\,$ Thus $\,{\rm mod}\ 3\!:\ 2\color{#c00}{a^3}+7a\equiv 2\color{#c00}{a}+7a\equiv 9a\equiv 0.$

Or use little Fermat, or proceed by cases $\,a\not\equiv 0\,\Rightarrow\, a\equiv \pm1\,\Rightarrow\,a^2\equiv 1\pmod 3.$ Or use integrality of binomial coefficients $\ a^3-a\, =\, 3!{a+1\choose 3}\equiv 0\pmod 3.\,$ Or use induction to prove that $\,f(a)\,$ is constant mod $\,3\,$ since $\,f(a+1)-f(a) = 3(2a^2+2a+3)\equiv 0.$

$7\equiv 1 \pmod{3}$ therefore your original problem reduces to $a(2a^2+1)$ If $a\equiv 0 \pmod{3}$ then this is divisible by 3. Assume it is not, then either $a\equiv 1$ or $a\equiv 2 \pmod{3}$.

$a\equiv 1$:

$1(2*1^2+1)\equiv 2+1\equiv 0 \pmod{3}$

$a\equiv 2$:

$2(2*2^2+1)\equiv 2*(8+1) \equiv 18 \equiv 0 \pmod{3}$

Let $f(a)=a(2a^2+7)$. There are three cases: 1) $a$ is a multiple of $3, a=3n$ say, 2) $a=3n+1, n \in Z$ and 3) $a=3n+2, n \in Z$.

Case 1) is obviously true.

After simplification,

Case 2) $f(a)=3 (3 n+1) \left(6 n^2+4 n+3\right)$ (true) and finally

Case 3) $3 (3 n+2) \left(6 n^2+8 n+5\right)$ (also true).

Q.E.D.