Math Puzzle: Prime Numbers
Posted on August 12, 2008
Filed Under Math Puzzles |
Take any prime number, square it and subtract 1. The result will be divisible by 24. Can you explain why?
(Examples: 5^2-1 = 24. 7^2-1 = 24*2. 11^2-1 =24*5, etc … )
Comments
6 Responses to “Math Puzzle: Prime Numbers”
Leave a Reply
Ahem… Now here’s my logic…
This law holds true only for prime nos greater than 3…
Let us call the prime no. as A and 1 as B…
So,
(A^2-B^2)
=(A+B)(A-B)
i.e. (A+1)(A-1)
Hence, we find that every number (A^2-1) is a product of two consecutive even numbers…
For 5 they are 4 & 6, for 7 they are 6 & 8, while for 11 they are 10 and 12…
Now, as A is odd no, A+1 and A-1 are consecutive even nos. Hence, both these nos are even…
Also, one of the nos will always be divisible by 3. This is because in (A-1), (A) & (A+1), A is prime and these are three consecutive nos. Plus this no. is divisible by 2 (as shown above). So, this no. is divisible by 6…
Similarly, at least one of the two nos is a multiple of 4. This is because in (A-1), (A) & (A+1), A is prime and (A-1) and (A+1) are even nos. To make a set of 4 consecutive nos, we can rope in (A-2) or (A+2). But these two nos will always be odd. So, either A-1 or A+1 are divisible by 4…
Thus the product will also be divisible by 4 * 6 i.e. 24…
Hence, every (A^2-1) results in a multiple of 24!
Hurrah! I was da first one to get da answer…
Such an easy puzzle…
That seems like an almost perfect solution.
In your deductions you’ve proved one of the numbers in (A+1)(A-1) is divisible by 6 and the other by 4. This however does not include the scenario in which just ONE of the numbers is divisible by both 6 and 4, which means it only needs to be divisible by 12 - ergo no cigar quite yet, unless I’ve missed out on a detail.
You’re only missing one step: showing p^2-1 divides by 8, not 4.
A straightforward proof would involve writing p (which is odd) as p=2*n+1, such that:
p^2 - 1 = (2n+1)^2 - 1 = 4n^2 + 4n = 4n(n+1)
Now it is obvious p^2-1 is divisible by 4. Furthermore, n(n+1) must be divisible by 2 since one of the numbers n, n+1 must be even (they are consecutive - if n is odd then n+1 is even, if n is even then n+1 is odd). Hence 4n(n+1) is divisible by 8.
This completes the proof :).
well here is my logic……
every prime number can be represented at (6k +/- 1).
so prime^2-1 = (6k +/- 1)^2 - 1 = 36k^2 +/- 12k=12k(3k +/- 1)……hence divisible by 24 whether k even or odd
Nice!
(Note for readers: using 6k+/-1 is just another way of saying your prime number doesn’t divide by 3.)
Nobody seems to have noticed that the result is false for p=2 or p=3.
The proof relies simply on the fact that
p^2-1=(p-1)(p+1).
Those two number s are both even (that assumes p is not 2) and one of them is divisible by 3 (that assumes p is not 3).
So the result is true for any prime number greater or equal to 5.
I forgot to mention that of the two even numbers one of them is divisible by 4 and hence the product is divisible by 8.