Add all proper divisors of the number and compare the sum with the original number.

Start the sum with 1 for numbers greater than 1. Then check possible divisors up to the square root of n. When a divisor is found, add both the divisor and its paired divisor, making sure not to add the same value twice for a perfect square.

Pseudocode:

function isPerfectNumber(n):
    if n <= 1:
        return false
    sum = 1
    divisor = 2
    while divisor * divisor <= n:
        if n % divisor == 0:
            sum = sum + divisor
            pairedDivisor = n / divisor
            if pairedDivisor != divisor:
                sum = sum + pairedDivisor
        divisor++
    return sum == n