Double Squares

Facebook Hacker Cup 2011の問題らしい。
そんなのあるんだ。

入力値が、２つの数値をそれぞれ２乗した値の和になる組み合わせを求める。
入力値が10の場合は、10 = 3^2 + 1^2 なので1通り。
1行目は、ケースの数です。（これに気づかず嵌ってしまった）

正答率60%

((288.0/(288+189))*100).round

設問

Credits: This challenge appeared in the Facebook Hacker Cup 2011.

A double-square number is an integer X which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because 10 = 3^2 + 1^2. Your task in this problem is, given X, determine the number of ways in which it can be written as the sum of two squares. For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2 + 3^2.
NOTE: Do NOT attempt a brute force approach. It will not work. The following constraints hold:
0 <= X <= 2147483647
1 <= N <= 100

Input sample:

You should first read an integer N, the number of test cases. The next N lines will contain N values of X.

5
10
25
3
0
1

Output sample:

e.g.

1
2
0
1
1

やってみた

#!/usr/bin/env ruby

def sqrt_count(n)
  s = Math.sqrt(n)
  (0..s).inject(0) do |res, i|
    sq = Math.sqrt(n - i**2)
    res+=1 if (sq >= i) and (sq == sq.to_i)
    res
  end
end

def header_skip_reader(file)
  f = open(file)
  f.gets
  f.lines{ |line| yield(line) }
  f.close
end

header_skip_reader(ARGV[0]) do |line|
  puts sqrt_count(line.to_i)
end

header_skip_readerはおまけ。
1行読み飛ばしてファイル操作するのって、open/gets/close以外で無いんだろうか。