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 April 30th, 2010, 03:23 PM #1 Newbie   Joined: Mar 2010 Posts: 13 Thanks: 0 Primes -- division Hello! Determine all primes $x$ such that $x+8$ is divisible by $\lfloor \sqrt{x} \rfloor$ .
 May 1st, 2010, 06:32 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 The only even prime (which is 2) has the property described. Suppose some odd prime p has the property, so that p + 8 has such a divisor n. Let q = (p + 8)/n. Clearly q > n, and q and n are both odd (else p = nq - 8 would be even). If q = n + 2, p = nq - 8 = n(n + 2) - 8 = (n - 2)(n + 4), but p is a prime, and n = 3 and p = 7 contradicts n² < p, so there is no such solution. If q = n + 4, p = nq - 8 = n(n + 4) - 8 = (n + 1)² + 2n - 9. Since p < (n + 1)², n < 5, and so n = 3 and p = 13. If q = n + 2r (where r ? 3), p = nq - 8 = n(n + 2r) - 8 = (n + 1)² + (2r - 2)n - 9, so n = 1, r ? 5 and p = 2r - 7, and so r = 5 and p = 3. Hence 2, 3 and 13 are the only primes with the specified property.
 May 1st, 2010, 06:43 PM #3 Member   Joined: Apr 2010 Posts: 34 Thanks: 0 Re: Primes -- division Notice that you have to take a square root of a prime number. You agree with these statements: 1. If there is an answer, it must be a prime number, and when you plug it into (x+/sqrt(x), you get a whole number. 2. Whole numbers are included in the set of rational numbers, 3. If x is a prime number, (x+/sqrt(x) is irrational when sqrt(x) is irrational. So if sqrt(x) is irrational, x is not a solution. 4. The square root of any number that is not a perfect square is irrational. So we need a number that is both prime and a perfect square. Since a prime number can not have more than two different factors (one and itself), any prime number whose factors are different has the maximum number of factors allowable and can not be a perfect square. Therefore, any prime number whose 2 factors are different will not work. The only prime left is 1, and it satisfies all the conditions.
 May 1st, 2010, 06:49 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Primes -- division Uh, WHAT??? You must not understand the floor function. That's ok; I don't know how to type it! And 1 is not prime. But besides that, everything looks fine.
 May 1st, 2010, 06:51 PM #5 Member   Joined: Apr 2010 Posts: 34 Thanks: 0 Re: Primes -- division I thought that a prime number is any number that is only divisible by 1 and itself, but that is wrong.
May 1st, 2010, 06:55 PM   #6
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Re: Primes -- division

Quote:
 Originally Posted by pitoten I thought that a prime number is any number that is only divisible by 1 and itself.
...any number whose DISTINCT divisors are 1 and itself.
Other definitions simply state "An integer greater than 2,..."

And those brackets around the square root of x mean floor function (i.e. greatest integer)

 May 1st, 2010, 06:56 PM #7 Member   Joined: Apr 2010 Posts: 34 Thanks: 0 Re: Primes -- division ah, sorry about that
May 1st, 2010, 06:58 PM   #8
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Re: Primes -- division

Quote:
 Originally Posted by pitoten ah, sorry about that
It doesn't offend me!
Check out that function and then look back at skip's answer; it's sufficient and easy to understand.

 May 2nd, 2010, 02:44 AM #9 Newbie   Joined: Mar 2010 Posts: 13 Thanks: 0 Re: Primes -- division Thanks a lot for help.

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