Nice one from ISI 2009

Suppose P is any non square number

Its easy to see that X_{(p - [√p]}) = P

All integers can be expressed as (p - [√p])

Let (p - [√p]) = n

so X_n = n + [√p] = n + <n>

As [√p] = <√ (p - [√p]) > hence proved

Proof of the property
[√p] = <√ (p - [√p]) >

As P is a non square number N²<P<(P+1)²

so [√p] = N

we have to prove <√ (p - [√p]) > = N

P can be written as N² + K where 1≤K≤2N

so we have to prove <√(N² + K - N)> = N

This is true because for N+1≤K≤2N we have N<√(N² + K - N)<N + 0.5

and for 1≤K≤N-1 we have N- 0.5 <√(N² + K - N)<N

for K = N its verified that <√(N² + K - N)> = N

So proved

Or maybe we will present it in a such a posh manner it will look better :D.

First lets convert the <> to something we are familiar with.

note that

<x> = \left[x+\frac{1}{2} \right]

So, we look at the sequence

a_n = n +\left[\sqrt n+\frac{1}{2} \right]

Now let n=m^2+k; 0 \le k \le 2m

Its easy to prove that

m^2<m^2+m < \left(m+\frac{1}{2}\right)^2<m^2+m+1<(m+1)^2

That means

\left[\sqrt {m^2+k} + \frac{1}{2} \right] = m ; 0 \le k \le m^2+m

and

\left[\sqrt {m^2+k} + \frac{1}{2} \right] = m+1 ; m^2+m+1 \le k \le 2m

With this we easily see that the numbers

(m^2-m+1,m^2-m,..., m^2,...,m^2+m)

get mapped to

((m-1)^2+1,(m-1)^2+2,....,m^2+2m)-\left{m^2 \right}

Thus every natural number is obtained and the squares are missed

sorry, we seem to have started posting together