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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 1 year, 5 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

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

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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