Prove that $ j^4+11j^2-12$ is divisible by $3$. $j$ is a natural number $j$ is a natural number. I know prove $j^4+11j^2-12$ is divisible by $4$, but I realy don't know how to prove, that $j^4+11j^2-12$ is divisible by $3$.
 A: Hint: You can factor $j^4+11j^2-12 = (j^2+12)(j^2-1) = (j^2+12)(j-1)(j+1)$. 
Now, break the proof into $3$ cases: 
(1) $j$ is a multiple of $3$
(2) $j$ is one more than a multiple of $3$
(3) $j$ is one less than a multiple of $3$.
A: If this looks intimidating to factor, we can look at it $\pmod{3}$ first:
$$j^4+11j^2-12\pmod{3} \equiv j^4-j^2\pmod{3}\equiv j^2(j^2-1) \pmod{3}\equiv j^2(j+1)(j-1)$$
For any number $j$, we have that $j\pmod{3}\in\{0,1,2\}$, so this polynomial will be $0\pmod{3}$.  This means that $3$ divides it.
A: Write as $j^4+11j^2-12=j^4-j^2 + 12(j^2-1)=j^2(j-1)(j+1)+12(j^2-1)$ then note that the first term contains the product of $3$ consecutive numbers, one of which must be a multiple of $3$, and the second term is a multiple of $12$, thus a multiple of $3\,$.
A: Look at it mod 3. We have
$$
j^4 + 11j^2 - 12 \equiv j^4 - j^2 \pmod{3}.
$$
For $j \equiv 0, 1, -1 \pmod{3}$, we have $j^2 \equiv j^4 \pmod{3}$, so that
$$
j^4 - j^2 \equiv 0 \pmod{3}.
$$
A: Using repeated differences and Newton's interpolation formula  we get
$$
j^4+11j^2-12
=
-12 \binom{j}{0} + 12 \binom{j}{1} + 36 \binom{j}{2} + 36 \binom{j}{3} + 24 \binom{j}{4}
$$
and so $j^4+11j^2-12$ is actually always a mutiple of $12$.
A: $j^4+11j^2-12=(j^2+12)(j+1)(j-1)$


*

*If $j$ is of the form $3k\implies 3|(j^2+12)$

*If $j$ is of the form $3k+1\implies 3|(j-1)$

*If $j$ is of the form $3k+2\implies 3|(j+1)$

A: Consider the following cases:


*

*$j\equiv0\pmod3 \implies j^4+11j^2-12\equiv0^4+11\cdot0^2-12\equiv0-0-12\equiv0\pmod3$

*$j\equiv1\pmod3 \implies j^4+11j^2-12\equiv1^4+11\cdot1^2-12\equiv1+11-12\equiv0\pmod3$

*$j\equiv2\pmod3 \implies j^4+11j^2-12\equiv2^4+11\cdot2^2-12\equiv16+44-12\equiv0\pmod3$

