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Prove that for any integer \(n>1\), \(n^5 + n^4 + 1\) is not a prime number.

Note by Saran Balachandar 11 months, 3 weeks ago

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\(Adding\quad and\quad subtracting\quad { n }^{ 3 }\\ \\ { n }^{ 5 }+{ n }^{ 4 }+n^{ 3 }-{ n }^{ 3 }+1\\ =\quad { n }^{ 3 }({ n }^{ 2 }+n+1)\quad -\quad { n }^{ 3 }-{ n }^{ 2 }-n+1\\ { =\quad n }^{ 3 }({ n }^{ 2 }+n+1)-n({ n }^{ 2 }+n+1)+({ n }^{ 2 }+n+1)\\ =\quad ({ n }^{ 3 }-n+1)({ n }^{ 2 }+n+1)\\ The\quad original\quad expression\quad is\quad expressed\quad as\quad a\quad product\quad of\quad two\quad factors\quad \\ \Longrightarrow \quad It\quad is\quad not\quad a\quad prime\quad number\quad .\) – Saran Balachandar · 11 months, 3 weeks ago

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TopNewest\(Adding\quad and\quad subtracting\quad { n }^{ 3 }\\ \\ { n }^{ 5 }+{ n }^{ 4 }+n^{ 3 }-{ n }^{ 3 }+1\\ =\quad { n }^{ 3 }({ n }^{ 2 }+n+1)\quad -\quad { n }^{ 3 }-{ n }^{ 2 }-n+1\\ { =\quad n }^{ 3 }({ n }^{ 2 }+n+1)-n({ n }^{ 2 }+n+1)+({ n }^{ 2 }+n+1)\\ =\quad ({ n }^{ 3 }-n+1)({ n }^{ 2 }+n+1)\\ The\quad original\quad expression\quad is\quad expressed\quad as\quad a\quad product\quad of\quad two\quad factors\quad \\ \Longrightarrow \quad It\quad is\quad not\quad a\quad prime\quad number\quad .\) – Saran Balachandar · 11 months, 3 weeks ago

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