(1) A composite number a is a positive integer number that is greater than 1 and can be expressed as the product of two smaller positive integer numbers, say b and c. This definition restricts b and c to be greater than 1 and neither b or c can be equal to a but they can be equal to each other, if say a=4.

(2) A prime number p is a positive integer number that is greater than 1 and not a composite number, i.e. it cannot be expressed as the product of two smaller positive integer numbers.

(3) Now, 1 is neither a composite number or a prime number according to the definitions above (if I defined them correctly).

(4) A perfect number P is a positive integer number that equals to the sum of its positive integer divisors, excluding itself.

(5) (Definition of a perfect number extends point (3)) That is, 1 is not a composite number, a prime number, nor a perfect number.

Conclusion: P cannot be 1 nor p. All P are a, but not the other way around.

Is this conclusion correct? What worries me is number 1. We know that division by zero is undefined, therefore we cannot add 1 and 0 and say that 1 is a perfect number. Because if we add zero to 1, then it means that zero is a factor of 1 and we can divide by zero, but we know that we cannot divide by zero. So, 1 cannot be a perfect number.

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    $\begingroup$ The definitions are correct. $1$ is not perfect, prime, or composite. $0$ is not a divisor of $1$. $\endgroup$ Nov 2, 2019 at 19:21
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    $\begingroup$ "A composite number a is a positive integer..." - negative integer also exists... why? $\endgroup$
    – LAAE
    Nov 3, 2019 at 0:44
  • $\begingroup$ @usiro : What about complex numbers then? As far as I know, in number theory, mathematicians only concern themselves with positive numbers. But this is interesting, say that we include negative integers then. Which integer numbers are considered to be primes? 3 will have divisors 3, 1, -1, -3, therefore 3 is no longer a prime number with this extended definition of prime numbers. What about 2? Its divisors are 2, 1, -1, -2, also not a prime number now. What about 1? Its divisors are 1, -1 therefore with this new extended definition of prime numbers, 1 is the only prime number there is. $\endgroup$ Nov 3, 2019 at 1:12
  • $\begingroup$ I haven't thought about this before, i.e. negatives when dabbling with primes, so if someone who knows more can jump in and either correct or expand on this, that would be great. $\endgroup$ Nov 3, 2019 at 1:14
  • $\begingroup$ @DickArmstrong: Usually, when we investigate composites, we consider the set $\mathbb{N}$ of positive integers. $\endgroup$ Nov 14, 2019 at 12:43

1 Answer 1


As commented already by Don Thousand, the definitions are correct.

I think you are confusing the definition of perfect number with $k$-perfect (or multiperfect) number.

So let $\sigma(x)$ be the usual sum of (all) positive divisors of $x$, including $x$.

Perfect numbers, in the traditional sense of the word, are numbers $N$ that satisfy $\sigma(N) = 2N$ (since they ought to satisfy $\sigma(N) - N = N$, per your Definition (4)).

Multiperfect (or $k$-perfect) numbers $M$, on the other hand satisfy $\sigma(M) = kM$.

So the usual perfect numbers are just the $2$-perfect numbers.

Since $\sigma(1)=1$ (because $\sigma$ is multiplicative), then $1$ is $1$-perfect.

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    $\begingroup$ Let me prove the assertion that if $p$ is prime, then $p$ is not perfect. Suppose that $p$ is prime and perfect. Then $$2p=\sigma(p)=p+1$$ which implies that $p=1$ (contradicting the fact that $1$ is not prime). $\endgroup$ Nov 14, 2019 at 12:53
  • $\begingroup$ Interesting, yeah that wouldn't make sense. The contradiction was perfect, thank you! $\endgroup$ Nov 14, 2019 at 17:03
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    $\begingroup$ @DickArmstrong, even more, you can also show that prime powers are not perfect. To this end, assume to the contrary that $q^k$ is perfect. Then we obtain $$\frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq 2$$ since $q \geq 2$ implies that $1/q \leq 1/2$, which further means that $$\frac{q - 1}{q} = 1 - \frac{1}{q} \geq \frac{1}{2},$$ thereby proving the last inequality. By the chain of inequalities, we obtain $$\frac{\sigma(q^k)}{q^k} < \frac{q}{q - 1} \leq 2$$ which implies that $\sigma(q^k) < 2q^k$. (cont'd.) $\endgroup$ Nov 15, 2019 at 15:06
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    $\begingroup$ In fact, the preceding argument actually proves that all prime powers are deficient. Therefore, no prime power is perfect. $\endgroup$ Nov 15, 2019 at 15:06

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