Positive integer

A positive integer is an integer greater than zero.

A positive integer $A=pqr$ with nonzero integers $p,q,r$ satisfying $1/A=1/p+1/q+1/r$. Equivalently, $pq+pr+qr=1$. A useful parametrization is $A=n(n+d)(n+e)$ for positive integers $n,d,e$ with $de=n^2+1$, coming from the signed triple $(n,-n-d,-n-e)$.

A number $p^2{q}^3$ for distinct prime numbers $p,q$. The ordering of the two primes matters because their exponents differ. Unique prime factorization identifies the ordered pair from the resulting integer.

A positive integer whose sum with its digit reversal has only odd numerical digits, with leading zeros forbidden in both the original and reversed representations. For example, $36+63=99$. Studying the arithmetic carries of mirrored digit pairs allows whole lengths to be counted without enumerating all integers of that length.

A positive integer $n$ for which some division $n=dq+r$ with integers $d,q,r>0$ and $r<d$ has $d,q,r$ as consecutive terms of a geometric progression in some order. Writing a reduced common ratio as $a/b$ with $a>b>0$ and $\gcd (a,b)=1$, the three terms have the form $t{b}^2,tab,t{a}^2$ for a positive integer $t$.

In this positive integer context, a power of two is $2^k$ for a nonnegative integer $k$, giving $1,2,4,8,\dots$. Allowing negative or real exponents gives a broader meaning of exponentiation, which is not intended in the integer divisibility criteria here.

A Mersenne number has the form $M_k=2^k-1$ for a positive integer $k$. Its binary representation consists of $k$ ones. For example, $M_4=15$ is composite, while $M_5=31$ is prime. If $k=ab$ with $a,b>1$, the finite geometric series factorizationhas two factors greater than one. Consequently a Mersenne prime requires a prime exponent, but $M_{11}=23\cdot 89$ shows that this condition alone is insufficient.

A Mersenne prime is a Mersenne number that is a prime number. It is the prime factor required by the Euclid-Euler theorem; $3,7,31,127$ generate $6,28,496,8128$. The family is named after Marin Mersenne.

A perfect number is a positive integer equal to the sum of all its positive proper divisors. The divisor itself is excluded: including it gives the equivalent condition $\sigma (n)=2n$ for the sum-of-divisors function. The smallest examples are $6=1+2+3$ and $28=1+2+4+7+14$.

For comparison, $8$ has proper divisors $1,2,4$, whose sum is $7$: it is a deficient number. The proper divisors of $12$ are $1,2,3,4,6$, whose sum is $16$: it is an abundant number. The proper-divisor sum classifies these examples with the same rule, and gives $s(1)=0$, so one is not a perfect number.

The interesting question is how such balances can be constructed without testing every smaller integer as a possible divisor. The Euclid-Euler theorem answers it for even integers: a Mersenne prime $2^p-1$ gives $2^{p-1}(2^p-1)$, and every even perfect number arises this way. For $p=2,3,5,7$, this produces $6,28,496,8128$. The connection links divisor sums to primality testing for a particular family, while keeping the necessary hypotheses visible.^[ref]

Euclid's Elements already gives the construction in Book IX, Proposition 36: add a succession of powers of two, and if their sum is prime, multiply by the final power of two. The modern finite geometric series $1+2+\cdots +2^{p-1}=2^p-1$ makes the same pattern transparent. This original statement supplies the forward construction, rather than the converse classification of all even perfect numbers.^[ref]

An even integer $n>0$ is a perfect number if and only ifThis forces $p$ to be a prime number, as shown under Mersenne number. It does not say that every prime exponent works, nor classify possible odd integers.^[ref]

For the forward direction, put $q=2^p-1$ and use coprimality of $q$ and $2^{p-1}$. Multiplicativity of the sum-of-divisors function and the finite geometric series giveThe prime number hypothesis makes $\sigma (q)=q+1$; replacing it with a composite number changes this factor.

For a short converse outline, write $n=2^{a}m$ with $a\ge 1$ and $m$ an odd integer. The condition $\sigma (n)=2n$ givesHere $t$ is a positive integer and a proper divisor of $m$. If $t>1$, the distinct divisors $1,t,m$ alone exceed the required sum $m+t$. Thus $t=1$ and $\sigma (m)=m+1$, forcing $m$ to be prime and $m=2^{a+1}-1$.

The prime exponents $2,3,5,7,13$ give Mersenne primes and therefore perfect numbers. But $p=11$ gives $2^{11}-1=2047=23\cdot 89$. The resulting candidate $N=2^{10}\cdot 2047=2096128$ hasso $s(N)=2325392>N$. It is an abundant number, not a perfect number. Checking only that the exponent is prime would miss this obstruction.

A positive integer whose proper-divisor sum is smaller than itself; equivalently, its sum-of-divisors function satisfies $\sigma (n)<2n$. For example, $s(8)=7<8$; also $s(1)=0<1$.

A positive integer whose proper-divisor sum exceeds itself; equivalently, its sum-of-divisors function satisfies $\sigma (n)>2n$. For example, $s(12)=16>12$.