Can u prove it?

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

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Note that $n^{4} + 4 = (n^{4} + 4n^{2} + 4) - 4n^{2} = (n^{2} + 2)^{2} - (2n)^{2} = ((n^{2} + 2) - 2n)((n^{2} + 2) + 2n).$

For $n \gt 1$ both of these last two bracketed terms are $\gt 1,$ thus proving that $n^{4} + 4$ is composite for $n \gt 1.$

Brian Charlesworth - 6 years, 1 month ago

Perfect!!!..

naitik sanghavi - 6 years, 1 month ago

Sophie-Germain Identity.

$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$

$(a,b)=(n,1)$ gives $n^4+4=(n^2+2+2n)(n^2+2-2n)$ ,

which is composite for $n\ge 2$ , since $n^2+2-2n\ge 2$ .

mathh mathh - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$