10-14-2005, 07:41 PM
10-15-2005, 12:16 AM
Funny challenge!
If Mr Caldwell does enter you're lost..... :wink:
If Mr Caldwell does enter you're lost..... :wink:
10-15-2005, 12:32 AM
I have the answer, without much time spent either: there can be only one.
Why? Well, think about it:
4 (perfect square) - 1 = 3 (prime)
9 (perfect square) - 1 = 8 (not prime)
16 (perfect square) - 1 = 15 (not prime)
25 (perfect square) - 1 = 24 (not prime)
36 (perfect square) - 1 = 35 (not prime)
49 (perfect square) - 1 = 48 (not prime)
64 (perfect square) - 1 = 63 (not prime)
Do you see a pattern here? The results increment like this:
a(n) = a(n-1) + (a(n-1) - a(n-2)) + 2
This means that the current result is found by adding to the difference between the previous two results. In other words:
8 + (8 - 3) + 2 = 8 + (5) + 2 = 15
15 + (15 - 8) + 2 = 15 + (7) + 2 = 24
24 + (24 - 15) + 2 = 24 + (9) + 2 = 35
Note the increments shown in parentheses. They have a difference of 2 (because of the +2 at the end).
A little math lesson for people who hate it.
Because this keeps happening (and rendering non-prime numbers), the only possible prime number that fits your criterion is the number 3.
Why? Well, think about it:
4 (perfect square) - 1 = 3 (prime)
9 (perfect square) - 1 = 8 (not prime)
16 (perfect square) - 1 = 15 (not prime)
25 (perfect square) - 1 = 24 (not prime)
36 (perfect square) - 1 = 35 (not prime)
49 (perfect square) - 1 = 48 (not prime)
64 (perfect square) - 1 = 63 (not prime)
Do you see a pattern here? The results increment like this:
a(n) = a(n-1) + (a(n-1) - a(n-2)) + 2
This means that the current result is found by adding to the difference between the previous two results. In other words:
8 + (8 - 3) + 2 = 8 + (5) + 2 = 15
15 + (15 - 8) + 2 = 15 + (7) + 2 = 24
24 + (24 - 15) + 2 = 24 + (9) + 2 = 35
Note the increments shown in parentheses. They have a difference of 2 (because of the +2 at the end).
A little math lesson for people who hate it.
Because this keeps happening (and rendering non-prime numbers), the only possible prime number that fits your criterion is the number 3.
10-15-2005, 12:39 AM
Quote:Funny challenge!
If Mr Caldwell does enter you're lost..... :wink:
I don't know Mr. Caldwell, but you had better post your code before he beats you to it!
10-15-2005, 12:48 AM
Quote:I have the answer, without much time spent either: there can be only one.
Why? Well, think about it:
4 (perfect square) - 1 = 3 (prime)
9 (perfect square) - 1 = 8 (not prime)
16 (perfect square) - 1 = 15 (not prime)
25 (perfect square) - 1 = 24 (not prime)
36 (perfect square) - 1 = 35 (not prime)
49 (perfect square) - 1 = 48 (not prime)
64 (perfect square) - 1 = 63 (not prime)
Do you see a pattern here? The results increment like this:
a(n) = a(n-1) + (a(n-1) - a(n-2)) + 2
This means that the current result is found by adding to the difference between the previous two results. In other words:
8 + (8 - 3) + 2 = 8 + (5) + 2 = 15
15 + (15 - 8) + 2 = 15 + (7) + 2 = 24
24 + (24 - 15) + 2 = 24 + (9) + 2 = 35
Note the increments shown in parentheses. They have a difference of 2 (because of the +2 at the end).
A little math lesson for people who hate it.
Because this keeps happening (and rendering non-prime numbers), the only possible prime number that fits your criterion is the number 3.
It can be made even simpler: Your looking for a prime number of the form N^2 -1. This can always be factored into (N+1)*(N-1).
In the case where N=2, one of the factors is 1, the other one happens to be prime (3).
I was hoping someone would start writing code, but no one bit.
:bounce:
10-15-2005, 01:50 AM
Mr Caldwell runs the Prime Page http://primes.utm.edu/
You can find there 3 is the only prime that fits the criterion.
I modified some code to do the search, but as it did'nt find anything, I checked the Prime Page....
I would give the prize to rpgfan, who actually thinked at the problem....
You can find there 3 is the only prime that fits the criterion.
I modified some code to do the search, but as it did'nt find anything, I checked the Prime Page....
I would give the prize to rpgfan, who actually thinked at the problem....
10-15-2005, 03:08 AM
Quote:I would give the prize to rpgfan, who actually thinked at the problem....No, I'm just a maths geek. Prime numbers seem to be a topic for discussion within the QB community right now.