In 1949 H.-E.Richert proved (1) that every positive integer typeset structure is a sum of distinct primes. For more information please look at (2), and (3).

However, if you consider $ n = 11 $, you will see that you can't express it as a sum of distinct prime numbers.

Is there an error in Richert’s theorem?


(1): Richert, H. E. (1949). Über Zerfällungen in ungleiche Primzahlen. (German). Math. Z., 52, 342-343.

(2): http://www.cs.cas.cz/portal/AlgoMath/NumberTheory/AdditiveNumberTheory/RichertTheorem.htm

(3): http://www.jstor.org/stable/2319891?seq=1#page_scan_tab_contents

  • 6
    $\begingroup$ $11$ is a sum of distinct primes, it is already prime. Richert's theorem allows for the "sum" to just have a single term. $\endgroup$
    – lulu
    Aug 6, 2015 at 17:18
  • 1
    $\begingroup$ Just as lulu pointed out, the theorem doesn't break down for n=11 if you allow for the "sum" to just have a single term. However, if you allow for the sum to contain only 1 term, then shouldn't this theorem work for n>1 ? $\endgroup$ Aug 6, 2015 at 17:30
  • 1
    $\begingroup$ +1 from me, a misguided but valid question is a good question! $\endgroup$
    – Alec Teal
    Aug 6, 2015 at 17:32
  • 1
    $\begingroup$ Possibly of interest: If you are interested in the question, I'd recommend going through the proof in detail. It's not that bad. I expect, but am not sure, that there is some cutoff beyond which you can require "sums" to have at least $2$ terms. After all, it is hard to imagine that $p-2$ has a $2$ in the sum AND $p-3$ has a $3$ in the sum, and so on. Might not be hard to establish the result. $\endgroup$
    – lulu
    Aug 6, 2015 at 17:33
  • 2
    $\begingroup$ To the earlier comment, the theorem is truly false for $4$ and $6$. There is no getting around that. $\endgroup$
    – lulu
    Aug 6, 2015 at 17:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.