# Find the largest number!

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

- 2 years, 5 months ago