# Relations between composite numbers, prime numbers, and perfect numbers.

(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.

• The definitions are correct. $1$ is not perfect, prime, or composite. $0$ is not a divisor of $1$. Nov 2, 2019 at 19:21
• "A composite number a is a positive integer..." - negative integer also exists... why?
– LAAE
Nov 3, 2019 at 0:44
• @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. Nov 3, 2019 at 1:12
• 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. Nov 3, 2019 at 1:14
• @DickArmstrong: Usually, when we investigate composites, we consider the set $\mathbb{N}$ of positive integers. Nov 14, 2019 at 12:43

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.

• 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). Nov 14, 2019 at 12:53
• Interesting, yeah that wouldn't make sense. The contradiction was perfect, thank you! Nov 14, 2019 at 17:03
• @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.) Nov 15, 2019 at 15:06
• In fact, the preceding argument actually proves that all prime powers are deficient. Therefore, no prime power is perfect. Nov 15, 2019 at 15:06