# Can u prove it?

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

Note by Naitik Sanghavi
3 years, 1 month ago

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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 $$n \gt 1.$$

- 3 years, 1 month ago

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Perfect!!!..

- 3 years, 1 month ago

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

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

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

- 3 years, 1 month ago

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