When does ⌊√2015(n−1)⌋=⌊√2015n⌋\left\lfloor\sqrt{2015(n-1)}\right\rfloor = \left\lfloor\sqrt{2015n}\right\rfloor hold?

For how many integers 1n2015 does the following equation hold? 2015(n1)=2015n

I have been struggling with this simple-looking problem for a while. What I have done so far:

If n504, we can make use of 2015(n1)<2015n1 to write
and hence the equation does not hold. For n>504, I get stuck. I wrote a MATLAB code to find such n that satisfies the equation. The first few values of n are 544,565,581,595, but I can't find a pattern. Can you please give me a hint?

PS. The problem comes from a 27th Chilean 2015-16 mathematical olympiad.


Edit: Directly solving OP's question (using the same idea as below):

2015n2015(n1)1 is true for n504 and false for n503

So when n=1,2,,503 we have 2015n takes only distinct values (as the difference between two consecutive values is greater than 1), for a total of 503 values.

For n=504,505,,2015, we have 2015n does not skip any values (as the difference between two consecutive values is less than 1), so 2015n takes all the values between 2015504=1007 and 20152015=2015, a total of 20151007+1=1009 values.

So when 1n2015, we have that 2015n takes 1512 distinct values; the non-distinct values duplicating the previous values, since they are increasing with n.

So the number of solutions of the equation is 20151512=503

The original problem in the link asks to determine the number of different values of n22015, for 1n2015. I think that your equation is equivalent to the number of duplicates (I will verify when I get a chance)

The solution is to take the difference of two consecutive terms and compare it with 1:


As long as n1008, no value will be skipped, so all values between 0 and 100822015=504 will be taken (505 values).

When n>1008 each term will generate a new value, So we have 20151008+1=1008 new values.

The total is 505+1008=1513 distinct values

The number of duplicates is 20161513=503. So for 502 values your equation holds.

Source : Link , Question Author : Amir Parvardi , Answer Author : Momo

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