Find the least $N$ so there is no square 
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000 \cdot N$ contains no square of an integer.

Let $x^2$ appear before $1000N$ so:
$(x+1)^2 - x^2 > 1000 \implies x \ge 500$
Let $A = [1000N, 1000(N+1)]$
So I let $x=500$ then, $x^2 = 250000$, obviously this is impossible since the set has a square already.
So now I need to set some number, $k^2 \ge 501^2 \equiv 0 \pmod{1000}$
So I need to solve the quadratic residue:
$k^2 \equiv 0 \pmod{1000}$ with constraints for $k$. 
Lets see: $500^2 = 250000$ and $501^2 = 251001$ and $502^2 = 252004$
$501^2 = 500^2 + 1000 + \sum_{k=1}^{1} 2k - 1$
$502^2 = 500^2 + 2(1000) + \sum_{k=1}^{2} 2k - 1$
The next $n$th square after $500$ is generalized:
$p = 500^2 + n(1000) + \sum_{k=1}^{n} 2k - 1$
Take $p \pmod{1000}$ 
$p \equiv n + \sum_{k=1}^{n} 2k - 1 \pmod{1000}$
$p \equiv n + n^2 \pmod{1000}$
$n(n+1) \equiv 0 \pmod{1000}$
$n \equiv 0$ and $n \equiv -1 \pmod{1000}$
But this method doesnt work properly. The final answer is: $N = 282$.
Please some hints only.

HINT : 
As you wrote, we have $x\ge 500$.
Observe that 
$$501^2=251001,\quad 502^2=252004,\quad 503^2=253009,\cdots$$
$$251\times 1000=251000,\quad 252\times 1000=252000,\quad 253\times 1000=253000,\cdots$$
Now we need to have
$$(500+a)^2-(250+a)\times 1000\gt 1000\Rightarrow  a\gt 31$$
where $N=250+a$.
