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Yearly Division 2 (Posted on 2015-08-06) |
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Find all positive integers n such that n+2014 divides n2 + 2014 and n+2015 divides n2 + 2015.
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Submitted by K Sengupta
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Solution:
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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.
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