Prove that $ j^4+11j^2-12$ is divisible by $3$. $j$ is a natural number

Question

$j$ 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 $3$.

Answer 1

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 $3$.

Answer 2

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.

Answer 3

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 $3\,$.

Answer 4

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 $$ j^4 - j^2 \equiv 0 \pmod{3}. $$

Answer 5

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 $12$.

Answer 6

$j^4+11j^2-12=(j^2+12)(j+1)(j-1)$

1. If $j$ is of the form $3k\implies 3|(j^2+12)$
2. If $j$ is of the form $3k+1\implies 3|(j-1)$
3. If $j$ is of the form $3k+2\implies 3|(j+1)$

Answer 7

Consider the following cases:

• $j\equiv0\pmod3 \implies j^4+11j^2-12\equiv0^4+11\cdot0^2-12\equiv0-0-12\equiv0\pmod3$
• $j\equiv1\pmod3 \implies j^4+11j^2-12\equiv1^4+11\cdot1^2-12\equiv1+11-12\equiv0\pmod3$
• $j\equiv2\pmod3 \implies j^4+11j^2-12\equiv2^4+11\cdot2^2-12\equiv16+44-12\equiv0\pmod3$