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Find the largest integer value of \(n\) such that \(n + 2015 \) divides \(n^{2015} + 1 \).

Note by Rony Phong 2 years, 5 months ago

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Note that \( a|b \) means that \( a \) divides \( b \)

Using the fact that \( a+b | a^n + b^n \) for odd \( n \), we have \( (n + 2015) | (n^{2015} + (2015)^{2015}) \).

So if \( (n + 2015) | (n^{2015} + 1) \), then \( (n + 2015) | (n^{2015} + (2015)^{2015}) - (n^{2015} + 1) = (2015)^{2015} - 1 \)

Now, \( a|b \) implies that \( a \leq b \). So we have \( n + 2015 \leq (2015)^{2015} - 1 \) or \( n \leq (2015)^{2015} - 2016 \)

Thus the maximum value of \( n = (2015)^{2015} - 2016 \)

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`*italics*`

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italics`**bold**`

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

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

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`\sum_{i=1}^3`

`\sin \theta`

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## Comments

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TopNewestNote that \( a|b \) means that \( a \) divides \( b \)

Using the fact that \( a+b | a^n + b^n \) for odd \( n \), we have \( (n + 2015) | (n^{2015} + (2015)^{2015}) \).

So if \( (n + 2015) | (n^{2015} + 1) \), then \( (n + 2015) | (n^{2015} + (2015)^{2015}) - (n^{2015} + 1) = (2015)^{2015} - 1 \)

Now, \( a|b \) implies that \( a \leq b \). So we have \( n + 2015 \leq (2015)^{2015} - 1 \) or \( n \leq (2015)^{2015} - 2016 \)

Thus the maximum value of \( n = (2015)^{2015} - 2016 \)

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