It's a Number Theory, Divisor Problem

If we let \(f(n)\) denote the sum of all the positive divisors of the integer \(n\),
how many integers i exist such that \(1 \le i \le 2010\) and \(f(i) = 1 + \sqrt{i} + i\)?

\(\small{ \begin{array}{|r|r|r|r|l|} \hline n & \text{prime number} & i=\text{prime number}^2 & \text{the sum of the positive divisors i } & f(i) = 1+\sqrt{i} + i \\ \hline 1 & 2 & 4 & 7 & f(2^2) = 1+2+2^2 = 7 \\ 2 & 3 & 9 & 13 & f(3^2) = 1+3+3^2 = 13 \\ 3 & 5 & 25 & 31 & f(5^2) = 1+5+5^2 = 31 \\ 4 & 7 & 49 & 57 & f(7^2) = 1+7+7^2 = 57 \\ 5 & 11 & 121 & 133 & f(11^2) = 1+11+11^2 = 133 \\ 6 & 13 & 169 & 183 & f(13^2) = 1+13+13^2 = 183 \\ 7 & 17 & 289 & 307 & f(17^2) = 1+17+17^2 = 307 \\ 8 & 19 & 361 & 381 & f(19^2) = 1+19+19^2 = 381 \\ 9 & 23 & 529 & 553 & f(23^2) = 1+23+23^2 = 553 \\ 10 & 29 & 841 & 871 & f(29^2) = 1+29+29^2 = 871 \\ 11 & 31 & 961 & 993 & f(31^2) = 1+31+31^2 = 993 \\ 12 & 37 &1396 & 1407 & f(37^2) = 1+37+37^2 = 1407 \\ 13 & 41 &1681 & 1723 & f(41^2) = 1+41+41^2 = 1723 \\ 14 & 43 &1849 & 1893 & f(43^2) = 1+43+43^2 = 1893 \\ \hline \end{array} }\)