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

Note by Naitik Sanghavi 2 years, 2 months 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 \(n \gt 1.\) – Brian Charlesworth · 2 years, 2 months ago

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@Brian Charlesworth – Perfect!!!.. – Naitik Sanghavi · 2 years, 2 months 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 · 2 years, 2 months ago

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TopNewestNote 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 · 2 years, 2 months ago

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– Naitik Sanghavi · 2 years, 2 months ago

Perfect!!!..Log in to reply

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 · 2 years, 2 months ago

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