Prove that $n^{4} + 4$ is a composite number for each natural number n is greater than 1.

Open in App

Solution

$n^{4} + 4 = (n^{4} + 4 n^{2} + 4) - 4 n^{2}$
$= (n^{2} + 2)^{2} - (2 n)^{2}$
$= (n^{2} + 2 + 2 n) (n^{2} + 2 - 2 n)$
This clearly shows that $n^{4} + 4$ is a composite number

Suggest Corrections

Similar questions

Prove n⁴+4 is a composite number for n>1

n^{4} + 4^{n}

n > 1.

Prove that n^4 +4 is a composite number for n>1.

n

5

n^{4} - 1

240

\int_{0}^{\infty} [n e^{- x}] d x = l n (\frac{n^{n}}{n!})

1

[\cdot]

Join BYJU'S Learning Program