The discussion focuses on solving the Euler Totient Puzzle by finding a number k such that phi(k+n) is congruent to 0 modulo 79 for n ranging from 0 to 79. The solution involves selecting 80 primes of the form 79n+1 and applying the Chinese Remainder Theorem to establish a system of congruences. Each prime p_n divides k+n, ensuring that phi(k+n) is divisible by 79. The participants acknowledge the necessity of proof to confirm the existence of such a k for every odd prime n.
PREREQUISITESMathematicians, number theorists, and students interested in modular arithmetic and prime number theory will benefit from this discussion.
Dodo said:Since there are infinite primes in any[*] arithmetic sequence, I'd choose 80 primes, call them p0 to p79, of the form 79n+1.
Next, use the Chinese Remainder Theorem to find a solution 'k' of the system of 80 congruences [itex]k \equiv -n \pmod {p_n}[/itex] with n=0,1,2,...79.
It follows that p0 divides k, p1 divides k+1, ... and p79 divides k+79. Then each (pn-1) will divide each phi(k+n); and since all primes were a multiple of 79 plus 1, each phi(k+n) will be divisible by 79, as required.
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Edit:
[*] Well, not any, but you get the idea.