Euler totient puzzle

Gaussianheart

Let n=79
phi is Euler totient

Can you find a number k such as :

phi(k)=0 mod 79
phi(k+1)=0 mod 79
phi(k+2)=0 mod 79
phi(k+3)=0 mod 79
.....
phi(k+78)=0 mod 79
phi(k+79)=0 mod 79

Good luck!

dodo

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 (p_n-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.

Gaussianheart

You are right.
Thanx for your solution.
Anyway it does not mean that it ALWAYS exists k for every n odd prime.
Proof is needed.