Ex-1 (a)
Using the formula
$2^{15} - 1$ = $2^{3*5} -1$
a = 5, b = 3
$2^b -1$ = $2^3 - 1$ = 7
and
$1 + 2^b + 2^{2b} + ... + 2^{(a-1)b}$ = $1 + 2^3 + 2^{2*3} + 2^{3*3} + 2^{4*3}$ = 4681
Ex-1 (b)
$2^{32767} - 1$ = $2^{7*4681} - 1$
b = 7, so
x = $2^7 - 1$ = 127
Ex-2
2 conjectures are:
$3^n - 2^n$ is prime for prime n
$3^n - 1$ is never prime except for n = 1
Ex-3(a)
2*3*5*7 + 1 = 211
Ex-3(b)
2*3*5*7*11 + 1 = 2311
Ex-4
x = (n+1)! + 2 = (5+1)! + 2 = 722
5 consecutive non-prime numbers are x, x + 1, x + 2, x + 3, x + 4 = 722,723,724,725,726
Ex-5
Use: If $2^n - 1$ is prime $\Rightarrow$ $2^{(n-1)}(2^n - 1)$ is perfect number
$2^5 - 1 = 31$ is prime $\Rightarrow$ $2^{(5-1)}(2^5 - 1)$ = 496 is perfect number
$2^7 - 1 = 127$ is prime $\Rightarrow$ $2^{(7-1)}(2^7 - 1)$ = 8128 is perfect number
Ex-6 There are no more triplet primes other than 3,5,7 where adjacent numbers in the list differ by 2.
Proof: I'll prove it by contradiction. Let say p, p+2, p+4 (p > 3)be such a triplet prime list.
Since one of the 3 consecutive numbers has to be divisible by 3, so
one out of p, p+1 or p+2 is divisible by 3. Since p and p+2 are primes by assumption so p+1 is divisible by 3. But then p+4 = (p+1)+3 and hence p+4 should be divisible by 3 and this is a contradiction.
Using the formula
$2^{15} - 1$ = $2^{3*5} -1$
a = 5, b = 3
$2^b -1$ = $2^3 - 1$ = 7
and
$1 + 2^b + 2^{2b} + ... + 2^{(a-1)b}$ = $1 + 2^3 + 2^{2*3} + 2^{3*3} + 2^{4*3}$ = 4681
Ex-1 (b)
$2^{32767} - 1$ = $2^{7*4681} - 1$
b = 7, so
x = $2^7 - 1$ = 127
Ex-2
| n | $3^n - 1$ | $3^n - 2^n$ |
| 1 | 2 | 1 |
| 2 | 8 | 5 |
| 3 | 26 | 19 |
| 4 | 80 | 65 |
| 5 | 242 | 211 |
| 6 | 728 | 665 |
| 7 | 2186 | 2059 |
| 8 | 6560 | 6305 |
| 9 | 19682 | 19171 |
| 10 | 59048 | 58025 |
2 conjectures are:
$3^n - 2^n$ is prime for prime n
$3^n - 1$ is never prime except for n = 1
Ex-3(a)
2*3*5*7 + 1 = 211
Ex-3(b)
2*3*5*7*11 + 1 = 2311
Ex-4
x = (n+1)! + 2 = (5+1)! + 2 = 722
5 consecutive non-prime numbers are x, x + 1, x + 2, x + 3, x + 4 = 722,723,724,725,726
Ex-5
Use: If $2^n - 1$ is prime $\Rightarrow$ $2^{(n-1)}(2^n - 1)$ is perfect number
$2^5 - 1 = 31$ is prime $\Rightarrow$ $2^{(5-1)}(2^5 - 1)$ = 496 is perfect number
$2^7 - 1 = 127$ is prime $\Rightarrow$ $2^{(7-1)}(2^7 - 1)$ = 8128 is perfect number
Ex-6 There are no more triplet primes other than 3,5,7 where adjacent numbers in the list differ by 2.
Proof: I'll prove it by contradiction. Let say p, p+2, p+4 (p > 3)be such a triplet prime list.
Since one of the 3 consecutive numbers has to be divisible by 3, so
one out of p, p+1 or p+2 is divisible by 3. Since p and p+2 are primes by assumption so p+1 is divisible by 3. But then p+4 = (p+1)+3 and hence p+4 should be divisible by 3 and this is a contradiction.
Just noticed - in your solution to Ex 6, 57 is not a prime - it's 19 x 3.
ReplyDeleteI was doing the same exercises, and came across this - http://en.wikipedia.org/wiki/Prime_triplet
thanks hrish. Now that I think about it again, I think another list of such triplet primes is not possible and I've added a quick proof for same.
ReplyDeletejust curious, what was your answer?
Himanshu, Initially I was under the impression that such a triplet might exist, and it might do so within the first 100 or 150 +ve integers. I actually went through the numbers and noticed the pattern in the wikipedia article I mention. A google search found it for me - so I did not actually prove it myself. Wish I had done that first :)
ReplyDeletedon't worry, there will be plenty of opportunities :)
ReplyDeleteBTW in Ex-3(b), the answer is not 111 because 111 is not a prime; It is divisible by 3.
ReplyDeletethanks @Niyaz, yes I missed 3 and 7. For the theorem to work I have to include all the primes from 2 to 11. Corrected that now.
ReplyDeleteHmm... the answer is correct, but this method is still not guaranteed to give you a prime number. (If it did, then it would be easy to find a larger prime number than the current largest prime number right? http://en.wikipedia.org/wiki/Largest_known_prime_number )
ReplyDeleteMy point was that 111 is not a prime number. 111 = 3 * 37, which gives us 37 (or 3) as our new prime number which the question asked us to find.
If you look at the proof of theorem 3, you can see that there are two cases in it. First when p1.p2.p3....pn + 1 is a prime and the second case where it is not a prime. I think Ex-3(a) wants us to use case 1 while Ex-3(b) wants us to use case 2.
@Niyaz: No, the method is guaranteed to give you a prime.
ReplyDelete2*3*5*....p_n + 1 will always be a prime number(however, ensure that you are taking all the primes smaller than and equal to p_n, you can't miss any). I guess, finding the largest prime is limited by the computational resources and not by theory. As theorem is all about proving that there are infinitely many primes.
E6 = 13# + 1 = 30031 = 59 x 509 is the first composite Euclid number, demonstrating that not all Euclid numbers are prime.
ReplyDeletehttp://en.wikipedia.org/wiki/Euclid_number
Thanks for correcting.. Yes, Agree, I was wrong.
ReplyDeleteThere exist infinite primes *does not* imply all euclid numbers are prime(which looked like true due to the way theorem was proved).
Stumbling across this in 2013 while working through the introduction; I think the above discussion should be mandatory reading after attempting the exercise!
ReplyDeleteIn fact, 3^n-2^n is a multiple of 5 whenever n is even. (Can be proved by induction.)
ReplyDeleteThe conjecture that '3^n-2^n is a prime number when n is prime' is not valid. This can be seen for n=7 where 3^7-2^7=2059=29*71.
ReplyDelete