Yearly Division 2 (Posted on 2015-08-06)

	Submitted by K Sengupta
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n+2014 divides n2 + 2014, and so: n+2014 divides n2 + 2014 – (n+2014) = n2 - n Similarly, n+2015 divides n2 - n Now, n+2014 and n+2015 are relatively prime and so it follows that: (n+2014)(2015) divides n2 - n .....(i) Since n is a positive integer it follows that: (n+2014)(2015) > n2 + 4029n > n2 - n Now, n^2 –n cannot be –ve as n is a +ve integer So, the only way that (i) will hold is for n2 - n to be precisely equal to zero. This gives n=1 (omitting n=0 since this is not permissible) Accordingly, n=1 constitutes the only solution to the problem under reference. *** Of course if n was a nonnegative integer (instead of a positive integer), then we would have n = 0 as an additional solution.