Suppose (in order to get a contradiction) that divides .
Let p be a prime divisor of . Then p must divide one or other of and . But this means that p divides two of the three terms in the relation and therefore it divides the third one as well. However, , and since p divides and it must also divide . Continuing in this way by "reverse induction", you see that p must divide , , ..., and eventually that p divides .
That contradiction shows more than was asked for, namely that none of the prime divisors of can divide or .
Edit: That proof doesn't quite work. It's okay unless p = 3 or 5. But those two primes will require some other argument. I don't have time to think about that just now.
Second edit: No real problem with p = 3 or 5. Suppose that p = 3. The recurrence relation says that if is a multiple of 3 then so is . The "reverse induction" procedure then tells us that either or is a multiple of 3, which is a contradiction.
In the same way, if p = 5 and is a multiple of 5 then the recurrence relation shows that is a multiple of 5. Continuing down the sequence to , we again get a contradiction.
Yes:
I made it also by induction, it is similar:
First I prove that every number from the sequence is not divisible by :
INDUCTION BASE: I see that doesn't divide , where .
Induction step: I will prove for arbitrary (integer)
that if is not divisible by therefore is also not divisible by .
I do similar with division by .
And then, also proving by induction I say:
Every three consecutive numbers in the sequence ( ) are relatively prime [we also know now that every number from the sequence is not divisible by and ]:.
INDUCTION BASE: Let - so are relatively prime.
Induction step: I will prove that if therefore . So, we see that are reletively prime - so is also relatively prime with - it is easy to deduce.
Taht's why there does not exist such (integer of course) that is divisor of
Can anybody check my proof, please - I am not a specialist