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In particular, the set of even integers that are not the sum of two primes has density zero. In 1951, Yuri Linnik proved the existence of a constant K such that every sufficiently large even number is the sum of two primes and at most K powers of 2. János Pintz and Imre Ruzsa found in 2020 that K = 8 works. [21]
The problem is NP-hard even when all input integers are positive (and the target-sum T is a part of the input). This can be proved by a direct reduction from 3SAT. [2] It can also be proved by reduction from 3-dimensional matching (3DM): [3] We are given an instance of 3DM, where the vertex sets are W, X, Y. Each set has n vertices.
The prime number theorem is obtained there in an equivalent form that the Cesàro sum of the values of the Liouville function is zero. The Liouville function is ( − 1 ) ω ( n ) {\displaystyle (-1)^{\omega (n)}} where ω ( n ) {\displaystyle \omega (n)} is the number of prime factors, with multiplicity, of the integer n {\displaystyle n} .
Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.
Lander, Parkin, and Selfridge conjecture: if the sum of -th powers of positive integers is equal to a different sum of -th powers of positive integers, then +. Lemoine's conjecture : all odd integers greater than 5 {\displaystyle 5} can be represented as the sum of an odd prime number and an even semiprime .
An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no factor p k, where prime and k is odd. In writing a number as a sum of two squares, it is allowed for one of the squares to be zero, or for both of them to be equal to each other, so all squares and all doubles of squares are ...
Subset sum problem, an algorithmic problem that can be used to find the shortest representation of a given number as a sum of powers; Pollock's conjectures; Sums of three cubes, discusses what numbers are the sum of three not necessarily positive cubes; Sums of four cubes problem, discusses whether every integer is the sum of four cubes of integers
For the avoidance of ambiguity, zero will always be a valid possible constituent of "sums of two squares", so for example every square of an integer is trivially expressible as the sum of two squares by setting one of them to be zero. 1. The product of two numbers, each of which is a sum of two squares, is itself a sum of two squares.