20210502, 10:03  #1 
Jul 2015
2·19 Posts 
May 2021

20210502, 12:30  #2 
"Walter S. Gisler"
Sep 2020
Switzerland
1111_{2} Posts 
I believe there are some mistakes in the problem statement, can anyone confirm this?
1. For every natural number , we have that for some , is equivalent to modulo (i.e. divides ). Shouldn't this be: For every natural number , we have that for some , is equivalent to modulo 2. one can prove that this sequence does not contain any primes by using the following easytoprove identity, which holds for any Fibonaccilike sequence: . And this one: ? 
20210502, 13:51  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3·23·149 Posts 
Isn't (1) the same either way?
My "argument" with them is F0=1, which is wrong. Everybody knows F0=0 (otherwise the "only prime indexes can be prime" won't hold). In fact, how I always remember it is that the 5th fibo number is 5 In fact, they say as much, further when affirming that F3=2 and F11=89. If F0=1, those would be false. Last fiddled with by LaurV on 20210502 at 14:06 
20210502, 14:44  #4 
"Walter S. Gisler"
Sep 2020
Switzerland
F_{16} Posts 
Ah, you are right. I had missed that F0 = 1. With that, 2 is indeed correct.
Regarding (1), I am pretty sure it isn't the same either way. m_k >= a_k, hence a_k mod m_k is either 0 or a_k , so with a set of triplets that is limited in size, a_k mod m_k couldn't be any natural number. 
20210502, 18:03  #5 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3·23·149 Posts 
"a (mod m)" is defined as the reminder (an integer number between 0 and m1, inclusive) that is left when you divide a to m, therefore, a\(\equiv\)b (mod m) is the same as b\(\equiv\)a (mod m), those are "equivalence classes", not numbers, it is just that we are lazy and don't write the "hats" (i.e. \(\hat{a} \equiv \hat{b}\) (mod m))

20210502, 18:43  #6 
"Walter S. Gisler"
Sep 2020
Switzerland
3·5 Posts 
Thanks, LaurV. 0scar also kindly explained it to me. This really confused me, but it is clear now.

20210502, 20:48  #7 
Sep 2017
2×73 Posts 
Also note that a_k can't be zero but can be m_k due to 1 <= a_k <= m_k

20210503, 05:52  #8 
Oct 2017
139 Posts 
Have A_0 and A_1 to be relatively prime? He didn't write it.
Otherwise A_0 = 4 and A_1 = 6 yielded a Fibonaccilike sequence without primes. But that cannot be the challenge. 
20210503, 14:08  #9 
Jan 2021
California
1014_{8} Posts 
That was just the start of it, you need to read the rest of the requirements of the series to see if the series you made meets the other requirements.

20210503, 18:12  #10  
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2^{2}×5^{2} Posts 
Quote:
GCD[A0,A1] must be = 1. I think that the webmaster did not write the relation of divisiblity between A0 and A1 because the question was to find a sequence of triplets with the conditions quoted 1 to 4 ”and the A0 A1 are simply to explain the goal of the question. Last fiddled with by Kebbaj on 20210503 at 18:18 

20210504, 12:50  #11 
Feb 2017
Nowhere
3^{4}·7·11 Posts 
The May 2021 Challenge seems quite similar to finding Riesel or Sierpinski numbers by means of "covering sets" of congruences.
The misstatement of the subscripts is the kind of oopsadaisy a number theorist might make The condition that the initial terms be relatively prime seems to have been assumed, but should have been stated. 
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