# the largest number that must be a divisor of the result.

• Jul 31st 2012, 12:12 AM
rcs
the largest number that must be a divisor of the result.
Select any prime number greater than 3. Square it and subtract 1. What is largest number that must be a divisor of the result?

My solution:

5^2- 1.= 25 - 1 = 24 the largest number of this 24 is 24x 10^+ infinity.... that's is how i got my answer.

can anybody give any suggestion or any assistance for this problem to be answered?

thanks
• Jul 31st 2012, 12:19 AM
Deveno
Re: the largest number that must be a divisor of the result.
p will be odd. therefore p2 will be odd. therefore p2-1 will be even.

the greatest divisor will be p2-1 itself. the next largest divisor will be (p2-1)/2.

(in your example p = 5. so p2 - 1 = 24. 24 is a divisor of 24. but, if you only want "proper divisors", the next largest divisor is 12).
• Jul 31st 2012, 11:12 AM
richard1234
Re: the largest number that must be a divisor of the result.
You want the largest integer that divides $p^2 - 1$, or $(p-1)(p+1)$.

We can show that 4 divides the above expression, as $p^2 \equiv 1 (\mod 4)$ for all odd p. Also, we can show that 3 divides the above expression, as 3 divides either p-1 or p+1. However we cannot conclude anything for larger prime divisors. Hence the answer is 12.
• Jul 31st 2012, 12:15 PM
HallsofIvy
Re: the largest number that must be a divisor of the result.
You are probably correct but that is assuming a problem different from the one initially posted. That said "Select any prime number greater than 3" so that we are working with a given prime number. 12 is the largest number that divides $p^2-1$ for all prime numbers larger than 3.
• Jul 31st 2012, 12:39 PM
richard1234
Re: the largest number that must be a divisor of the result.
Yes but that would mean the answer could be arbitrarily large. Also the problem says, "select any...," "largest prime number that must..."
• Jul 31st 2012, 05:33 PM
rcs
Re: the largest number that must be a divisor of the result.
thank you richard1234, Deveno and HallsOfIvy
• Jul 31st 2012, 11:14 PM
rcs
Re: the largest number that must be a divisor of the result.
how is it formally expressed in mathematical expression that "Select any prime number greater than 3. Square it and subtract 1. What is largest number that must be a divisor of the result?"

Thanks
• Jul 31st 2012, 11:21 PM
richard1234
Re: the largest number that must be a divisor of the result.
Quote:

Originally Posted by rcs
how is it formally expressed in mathematical expression that "Select any prime number greater than 3. Square it and subtract 1. What is largest number that must be a divisor of the result?"

Thanks

Something like $\max (\{k: k | p^2 - 1 \forall p \hspace{1 mm}prime, p \ge 3 \})$...I'm not entirely proficient at set-builder notation.
• Aug 1st 2012, 12:02 AM
rcs
Re: the largest number that must be a divisor of the result.
thank you sir
• Aug 1st 2012, 12:17 AM
rcs
Re: the largest number that must be a divisor of the result.
Quote:

Originally Posted by richard1234
You want the largest integer that divides $p^2 - 1$, or $(p-1)(p+1)$.

We can show that 4 divides the above expression, as $p^2 \equiv 1 (\mod 4)$ for all odd p. Also, we can show that 3 divides the above expression, as 3 divides either p-1 or p+1. However we cannot conclude anything for larger prime divisors. Hence the answer is 12.

Does 12 mean for all prime numbers greater than 3 is the largest number that will divides the result?

thank you
• Aug 1st 2012, 08:47 AM
richard1234
Re: the largest number that must be a divisor of the result.
Yes...LCM(4,3) = 12.