bonanova 85 Posted February 18, 2008 Report Share Posted February 18, 2008 Prove or disprove: If p is a prime number > 3, then p^{2}-1 is divisible by 24. Quote Link to post Share on other sites

0 Guest Posted February 18, 2008 Report Share Posted February 18, 2008 Prove or disprove: If p is a prime number > 3, then p^{2}-1 is divisible by 24. I love primes Things to note, p is a prime > 3 => 1. p is odd. 2. p is not divisible by any other number apart from 1 and itself => p is not divisible by 3. Now the solution: p^{2}-1 = (p+1)(p-1) => p^{2}-1 is product of 2 consecutive even numbers (from 1) Now product of 2 consecutive even number is always divisible by 8. (lets say 2n*(2n+2) = 4n(n+1)^{**} =4*(2m-1)*(2m-1+1) = 8m(2m-1) ) ^{**} - Note: n(n+1) is again a product of consecutive number and hence is divisible by 2 which is shown in the afore mentioned steps Also note that one of any 3 consecutive number is always divisible by 3, and since it cannot be p (from 2); it has to be either (p-1) or (p+1). Thus, p^{2}-1 is divisible by 8 and 3. Hence its divisible by 24! Quote Link to post Share on other sites

0 Guest Posted February 22, 2008 Report Share Posted February 22, 2008 First of all, any prime number squared is divisible by 24. If you are talking about the dividion resulting in an integer, then that's true too. Quote Link to post Share on other sites

0 Guest Posted February 22, 2008 Report Share Posted February 22, 2008 First of all, any prime number squared is divisible by 24. If you are talking about the dividion resulting in an integer, then that's true too. No. 51 .. No. The definition of "divisible" given by www.dictionary.com is: di·vis·i·ble /dɪˈvɪzəbəl/ Pronunciation Key - [di-viz-uh-buhl] –adjective 2. Mathematics. a. capable of being evenly divided, without remainder. b. of or pertaining to a group in which given any element and any integer, there is a second element that when raised to the integer equals the first element. Also, prime squared is not divisible by any other number apart from that prime number saquared, that prime number and 1. On the other hand, p^{2}-1 is exactly divisible by 24. I rest my case. Quote Link to post Share on other sites

0 bonanova 85 Posted February 22, 2008 Author Report Share Posted February 22, 2008 First of all, any prime number squared is divisible by 24. If you are talking about the dividion resulting in an integer, then that's true too. By definition of prime number the square of any prime number is divisible by itself, the prime number and 1. Quote Link to post Share on other sites

0 Guest Posted February 24, 2008 Report Share Posted February 24, 2008 p^{2}-1 = (p+1)(p-1) This is a remarkable little statement which is so simple and elegant that I can't believe I hadn't either seen it or discovered it before (I'm probably giving away my lack of math education, but that's ok). Thanks! Quote Link to post Share on other sites

0 Guest Posted February 25, 2008 Report Share Posted February 25, 2008 p is prime. p>3. 1. All prime numbers are odd, p is odd 2. p^2-1=(p-1)(p+1). 3. so p^2-1 is product of two consecutive even numbers. 4. product of two consecutive even numbers are divisible by 8. proof: let m be odd product=2m*(2m-2)=4(m(m-1)), as m is odd so m-1 is even and equals to 2n so, product=8*m*n-->divisible by 8. 5. p-1,p,p+1 are three consecutive numbers. So, one of them should be divisible by 3. p is prime, so either p-1 or p+1 falls into this category. 6. Any number divisible by both 3 and 8 is divisible by 24 Quote Link to post Share on other sites

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Prove or disprove:If p is a prime number > 3, then p

^{2}-1 is divisible by 24.## Link to post

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