Find the largest number!

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

#NumberTheory

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Comments

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 $n = (2015)^{2015} - 2016$

Siddhartha Srivastava - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$