I don't speak English well, but I have a serious problem.
Let . The sequence is defined by recursive equation for .
We must prove that this sequence contains only finitely many primes.
Thank you for our answers.
There is, however, a very easy answer. What's an easy test for divisibility by 3 based on a number's decimal digits?
EDIT: Sorry, I completely forgot what the question asked momentarily, and the above gives a reason for there being infinitely many composites. I'll have to revisit the question. (And I may not be able to find an answer.)
(By the way, does anyone know how to get strikethrough text formatting? I tried searching but couldn't find a way. So I made my wrong answer gray instead.)
Edit 2: If the right answer involves divisibility tests for small primes then this link ("Divisibility by prime numbers under 50" by Stu Savory) might be helpful (or might not).
use the fact that
You use the first fact to show that an inverse exists for and the second fact to show that the numerator is 0.
Actually this sort of reasoning outlines a possible line of attack for the n even case.
We want to find such that
thus if we can find a that satisfies
then we are done.
Following this logic, it would probably be best to find such that
I think the key is
From , is prime . we get
If , then we have the inequality
For to be prime , we must have
and which gives , a contradiction .
Since is finite , it is impossible to find a prime in for .
ps: Oh , I have just realized that is prime is unnecessary .
Forget about it
is an integer .
Therefore, the denominator is divisible by the numerator .
as the product of two integers that one is cancelled by the numerator and one remains .
Let or whatever you like .
After the division ,we obtain the quotient , if it is a prime , then either or is one , let so we have , a contradiction because we always have for so . Similar work on letting , we will have another contradiction so the quotient is not a prime for all .