From a blog whose name I have now forgotten, a math problem from a shabbos table:
Show that if p is a prime greater than or equal to 7 then p^4 - 1 is divisible by 240.
If p is a prime >= 7, then p is odd, therefore p = 2k+ 1 for some integer k >= 3. Thus,
p^4 - 1 = (p + 1)(p - 1)(p^2 + 1)
= (2k + 2)(2k)(4k^2 + 4k + 2)
= 8(k + 1)(k)(2k^2 + 2k + 1)
Since 240 = 8 * 2 * 3 * 5, it is now (necessary and) sufficient to show that (k + 1)(k)(2k^2 + 2k + 1) is divisible by 2 * 3 * 5.
(1) (k + 1)(k)(2k^2 + 2k + 1) is divisible by 2 since either k is even or k + 1 is even.
(2) (k + 1)(k)(2k^2 + 2k + 1) is divisible by 3 since one of
(a) k mod 3 = 0
(b) k mod 3 = 2 so that k + 1 mod 3 = 0
k mod 3 = 1 is impossible since if so then p = 2*(3m + 1) + 1 = 6m + 3 for some positive integer m and then p is not prime, contradicting our assumption.
(3) (k + 1)(k)(2k^2 + 2k + 1) is divisible by 5 since one of
(a) k mod 5 = 0
(b) k mod 5 = 4 so that k + 1 mod 5 = 0
(c) k mod 5 = 1, 2 or 3, so that 2k^2 + 2k + 1 mod 5 = 0.
Ta da! I'm sure there are more elegant solutions out there; feel free to offer one.
6 comments:
I'm not seeing (3)(c) when k mod 5 = 2.
And I thus add to the vast body of evidence that I can't multiply in my head.
However, if k mod 5 = 2 then
p = 2(5m + 2) + 1 = 10m + 5 and p is not prime, a contradiction.
Thanks :)
That is great. Well done!
Ah! That answers my question about why p must be >=7 (because if p = 5, then p is in fact 10m+5, so k mod 5 = 2, etc.).
I don't know if it was my post that started this all: http://floatingbear.blogspot.com/2007/05/shabbos-table-math.html#comments, but in any case there is a way to prove it for p>5, not just p>7. (not that i recall what that way is...)
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