In 1852, Chebyshev proved the Bertrand’s postulate:
For any integer , there always exists at least one prime number with .
Every integer can be expressed as a sum of distinct primes.
Consider the set of prime integers where . By Bertrand’s postulate we know that .
Next, we observe that, any integer between 7 and 19 can be written as a sum of distinct first 5 prime integers :
7 = 5+2; 8 = 5+3; 9 = 7+2; 10 = 5+3+2; 11 = 11; 12 = 7+5; 13 = 11+2; 14 = 7+5+2; 15 = 7+5+3; 16 = 11+5; 17 = 7+5+3+2; 18 = 11+7; 19 = 11+5+3
Hence we fix , , and to conclude that .
Let, . Then by the above observation we know that the elements of are the sum of distinct first prime integers.
Moreover, if the elements of can be written as the sum of distinct first prime integers, then the elements of can also be written as the sum of distinct first prime integers since
as a consequence of .
Hence inductively the result follows by considering , which contains all integers greater than , and contains only elements which are distinct sums of primes.
Exercise: Use Bertrand’s postulate to generalize the statement proved earlier: If and are natural numbers , then the sum
cannot be an integer.
[HINT: Look at the comment by Dan in the earlier post.]
 Turner, C. (2015) A theorem of Richert. Math.SE profile: https://math.stackexchange.com/users/37268/sharkos
 Sierpiński,W. (1988). Elementary theory of numbers. North-Holland Mathematical Library 31, pp. 137-153.