I could use a little guidance for this questions.
My thoughts, given the conditions n must be an even number, so n=2k where k > 3Let n be an integer greater than 6. Prove that if is n-1 and n+1 are both prime, then [tex]n^2(n^2+16) is divisible by 720. also is the converse true.
also i know that [tex] n^2 - 1 = a*b[tex] where a and b are prime and b=a+2
Could i have some help please, thanks in advance.
As well as being an even number, n has to be a multiple of 3 (otherwise one of the numbers n–1 and n+1 would have to be a multiple of 3. Therefore n is a multiple of 6 and so is a multiple of 36. Also, must be a multiple of 4.
So far, that tells us that is a multiple of 36×4=144. To get it as a multiple of 720 we only need an extra factor of 5, so we want to show that either n or must be a multiple of 5. One way to do this is to work mod 10. A prime number greater than 5 has to end in 1, 3, 7 or 9. So an even number with a prime on both sides of it has to end in 0, 2 or 8. If n ends in 0 then it already has a factor of 5 in it. if n ends in a 2 or an 8 then ends in a 4, so ends in a 0 ...
What about the converse? That had better not be true, otherwise there would be an easy way to construct an infinite number of prime pairs. So look for a counterexample (remembering that we started out by looking at n being a multiple of 6).
I was going to respond yesterday but I completely forget.
We want to show since it is equivalent to saying , , .
First we prove the case with . Since are primes (greater than 6) it means must be even because otherwise shall not be primes. But then is divisible by and is divisible by so their product is divisible by .
The next case is with . Now where by the division algorithm. It cannot be that because otherwise cannot be prime (since it is greater than 6). Similarly because then which is not prime. So is divisible by . Thus, is divisible by .
Finally the last case. By using the same argument as in the middle case we have meaning thus . Thus, is divisible by 5.
EDIT: Opalg beat me by 1 minute .
thank a lot to both of you for the help.
I'm very new to modular arithmetic, I just read this article nrich.maths.org :: Mathematics Enrichment :: Modular Arithmetic to give me a start. (can you suggest better articles?)
about the divisiblity of 5, I totally understand your mod10 argument Opalg, however ThePerfectHacker i don't understand how this works
because if k is odd then that means that n would also be odd but the conditions give rise to n being an even number, sorry i am a bit new to this, could you explain in more detail how the mod 5 works?
thank a lot for all your help,