Let $p_1, p_2, \dots, p_n$ be the first $n$ primes and $p_n\#$ be the primorial of $p_n$

Using the Chinese Remainder Theorem and Fermat's Little Theorem, the following, I believe, is true:

$$\sum_{i}^{n}\left(\frac{p_n\#}{p_i}\right)^{p_i-1} \equiv 1 \pmod {p_n\#} $$

Here's my reasoning:

Since $x^{p_i-1} \equiv 1 \pmod {p_i}$, it follows that the $\left(\frac{p_n\#}{p_i}\right)^{p_i-2}$ is the inverse of $\frac{p_n\#}{p_i}$ modulo $p_i$

So, it follows that there exists an $x$ such that:

$$\sum_{i}^{n}\left(\frac{p_n\#}{p_i}\right)^{p_i-1} = x{p_n\#} + 1$$

Is there a straight forward way to establish this expression without appeal to the Chinese Remainder Theorem or Fermat's Little Theorem. Does $x$ have any interesting properties?

  • 1
    $\begingroup$ Your proof is correct and beautiful. But I do not think, we can prove the claim without Fermat's little theorem. The chinese remainder theorem is not necessary : If the sum is denoted by $S$, the fact $p_i|S-1$ for $i=1,...,n$ easily implies $p_n\#|S-1$. The number $x$ seems to be virtually random. $\endgroup$ – Peter Oct 20 '15 at 18:05
  • $\begingroup$ For $n=7$, we have $$x=856803164036300304666023883076433130455364964554341823085399142787$$ $\endgroup$ – Peter Oct 20 '15 at 18:10
  • $\begingroup$ [3, 1; 11, 1; 19, 1; 576787, 1; 411019290856981, 1; 1490363591963507, 1; 3867618 412453392387965254789, 1] is its factorization. $\endgroup$ – Peter Oct 20 '15 at 18:12
  • $\begingroup$ Thanks Peter. I appreciate your analysis. $\endgroup$ – Larry Freeman Oct 20 '15 at 23:12

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.