How to find 5 repeated values in O(n) time?
The solution in fade2black's answer is the standard one, but it uses $O(n)$ space. You can improve this to $O(1)$ space as follows:


Let the array be $A[1],\ldots,A[n]$. For $d=1,\ldots,5$, compute $\sigma_d = \sum_{i=1}^n A[i]^d$.
Compute $\tau_d = \sigma_d - \sum_{i=1}^{n-5} i^d$ (you can use the well-known formulas to compute the latter sum in $O(1)$). Note that $\tau_d = m_1^d + \cdots + m_5^d$, where $m_1,\ldots,m_5$ are the repeated numbers.
Compute the polynomial $P(t) = (t-m_1)\cdots(t-m_5)$. The coefficients of this polynomial are symmetric functions of $m_1,\ldots,m_5$ which can be computed from $\tau_1,\ldots,\tau_5$ in $O(1)$.
Find all roots of the polynomial $P(t)$ by trying all $n-5$ possibilities.


This algorithm assumes the RAM machine model, in which basic arithmetic operations on $O(\log n)$-bit words take $O(1)$ time.



Another way to formulate this solution is along the following lines:


Calculate $x_1 = \sum_{i=1}^n A[i]$, and deduce $y_1 = m_1 + \cdots + m_5$ using the formula $y_1 = x_1 - \sum_{i=1}^{n-5} i$.
Calculate $x_2 = \sum_{1 \leq i &lt; j \leq} A[i] A[j]$ in $O(n)$ using the formula
$$
x_2 = (A[1]) A[2] + (A[1] + A[2]) A[3] + (A[1] + A[2] + A[3]) A[4] + \cdots + (A[1] + \cdots + A[n-1]) A[n].
$$
Deduce $y_2 = \sum_{1 \leq i &lt; j \leq 5} m_i m_j$ using the formula
$$
y_2 = x_2 - \sum_{1 \leq i &lt; j \leq n-5} ij - \left(\sum_{i=1}^{n-5} i\right) y_1.
$$
Calculate $x_3,x_4,x_5$ and deduce $y_3,y_4,y_5$ along similar lines.
The values of $y_1,\ldots,y_5$ are (up to sign) the coefficients of the polynomial $P(t)$ from the preceding solution.


This solution shows that if we replace 5 by $d$, then we get (I believe) a $O(d^2n)$ algorithm using $O(d^2)$ space, which performs $O(dn)$ arithmetic operations on integers of bit-length $O(d\log n)$, keeping at most $O(d)$ of these at any given time. (This requires careful analysis of the multiplications we perform, most of which involve one operand of length only $O(\log n)$.) It is conceivable that this can be improved to $O(dn)$ time and $O(d)$ space using modular arithmetic.