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n^4+4 prime number <==> n=1

Note by Mouataz Chadmi 4 years, 3 months ago

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\( n^4 + 4 = (n^2-2n+2)(n+2n+2)\) So the expression \( n^4+4\) can be factored for every \(n > 1\) and hence it can never be a prime for \(n>1\)

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That's not complete, you should mention that \(n^2-2n+2=1 \implies (n-1)^2=0 \implies n = 1\), and if \(n>1\), then both factors are bigger than \(1\), hence \(n^4+4\) is not prime.

Oh yeah, forgot that! That's why you're on level 5 and i'm on 4!

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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

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

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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}`

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TopNewest\( n^4 + 4 = (n^2-2n+2)(n+2n+2)\) So the expression \( n^4+4\) can be factored for every \(n > 1\) and hence it can never be a prime for \(n>1\)

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That's not complete, you should mention that \(n^2-2n+2=1 \implies (n-1)^2=0 \implies n = 1\), and if \(n>1\), then both factors are bigger than \(1\), hence \(n^4+4\) is not prime.

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Oh yeah, forgot that! That's why you're on level 5 and i'm on 4!

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