If $$n$$ is positive integer $$(n0)$$, show that $$n^4-3n^2+9$$ is composite number

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If $$n$$ is positive integer $$(n>0)$$, show that $$n^4-3n^2+9$$ is composite number

Uniteasy Admin · Accepted Answer

$$n^4-3n^2+9$$ 
$$=n^4+6n^2+9-9n^2$$
$$(n^2+3)^2 - (3n)^2$$
$$=(n^2+3n+3)(n^2-3n+3)$$
Since $$n>0$$,
$$n^2+3n+3>1$$ and $$n^2-3n+3>1$$
$$n^4-3n^2+9$$ is the product of two different integers that are greater than 1.
Therefore
It is a composite number

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