Question

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

Collected in the board: Number Theory

Steven Zheng posted 9 hours ago

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

Steven Zheng posted 9 hours ago

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