How to Check Whether a Number Is Perfect

A perfect number is a positive integer equal to the sum of its positive divisors other than itself. You can check a particular number by finding and adding those proper divisors. For larger numbers, pairing divisors or using the prime factorization avoids listing every divisor one by one.

The method works for any positive integer. It also clarifies a useful equivalent test: the sum of all positive divisors must be twice the number.

Definition and equivalent test

Write $s(n)$ for the sum of the proper positive divisors of $n$. Then $n$ is perfect precisely when

$$s(n)=n.$$

If $\sigma(n)$ denotes the sum of all positive divisors, including $n$, then $\sigma(n)=s(n)+n$. Therefore the same test can be written as

$$\sigma(n)=2n.$$

For example, the proper divisors of 6 are 1, 2, and 3. Their sum is 6, so 6 is perfect. The number 1 is not perfect: it has no proper positive divisors, so their sum is 0.

Check by pairing divisors

For a number $n$, each divisor $d$ is paired with $n/d$. One member of the pair is at most $\sqrt n$, and the other is at least $\sqrt n$. This means you only need to test possible divisors up to the square root.

Use this procedure:

1. Start the proper-divisor sum at 1 for $n>1$; 1 is always a proper divisor of such a number.
2. For each integer $d$ from 2 through $\lfloor\sqrt n\rfloor$, check whether $d$ divides $n$.
3. If it does, add both $d$ and $n/d$ to the sum.
4. If $n$ is a square and $d=\sqrt n$, add that divisor only once.
5. Compare the result with $n$.

For $n=28$, the divisors found from the test values are $2$ and $14$, and $4$ and $7$. Including 1 gives

$$1+2+14+4+7=28.$$

Thus 28 is perfect. The test did not need to check values above $\sqrt{28}$, because any larger proper divisor would have a matching divisor below the square root.

As a square-number example, take $n=36$. The proper divisors are $1,2,3,4,6,9,12,18$, whose sum is 55. When testing up to $\sqrt{36}=6$, the pair for $d=6$ is just $6$ itself; counting it twice would give the wrong sum. Since 55 is not 36, the number is not perfect.

Use the prime factorization

If you know the prime factorization

$$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},$$

you can calculate the sum of all divisors without listing them:

$$\sigma(n)=\prod_{i=1}^{k}(1+p_i+p_i^2+\cdots+p_i^{a_i}) =\prod_{i=1}^{k}\frac{p_i^{a_i+1}-1}{p_i-1}.$$

Then check whether $\sigma(n)=2n$. For 28, the factorization is $2^2\cdot7$, so

$$\sigma(28)=(1+2+4)(1+7)=7\cdot8=56=2\cdot28.$$

Therefore 28 is perfect. This formula is particularly convenient when a number has few prime factors, though finding the factorization can itself be difficult for very large numbers.

A shortcut for even perfect numbers

There is a special characterization of even perfect numbers: an even integer is perfect exactly when it has the form

$$2^{p-1}(2^p-1),$$

where both $p$ and the Mersenne number $2^p-1$ are prime. For instance, $p=3$ gives $2^2(2^3-1)=4\cdot7=28$. This shortcut is useful when checking candidates of that form. It does not settle whether any odd perfect numbers exist; that remains an open problem. The divisor-sum test still applies to any specific odd number you want to check.

Common mistakes

• Including $n$ among the proper divisors. The proper-divisor sum excludes the number itself. If you use $\sigma(n)$ instead, compare it with $2n$.
• Counting a square-root divisor twice. When $n$ is a square, its square root pairs with itself and should be added once.
• Checking divisors only up to $n/2$ without a reason. That method can work, but testing up to $\sqrt n$ and adding divisor pairs is usually faster.
• Assuming a familiar form proves perfection automatically. For the even-number characterization, verify that both $p$ and $2^p-1$ are prime.

For a single moderate-sized integer, the divisor-pair test is straightforward. For a number already factored into primes, the sigma formula is often shorter.