Difference between revisions of "Talk:Wilson's Theorem"
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Ok, I will be working on a few examples, and maybe some history to go along. | Ok, I will be working on a few examples, and maybe some history to go along. | ||
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| + | Solution needed for the first example. --[[User:Chess64|Chess64]] 17:58, 19 June 2006 (EDT) | ||
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| + | The solution to that first example, while maybe clear enough for a Mathlinks forum, seems a little difficult to understand, especially if you're just learning this theorem. How about: | ||
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| + | ''Solution:'' | ||
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| + | <math>(q!)^2 \equiv (2q)!(-1)^q \equiv (p-1)!(-1)^q \mod p </math> | ||
| + | since <math>(q+1)\cdot (q+2) \cdot ... \cdot (2q) \equiv (-q)\cdot (-q+1) \cdot ... \cdot (-1) \mod p</math> | ||
| + | so | ||
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| + | <math>(q!)^2+(-1)^q \equiv (-1)^{q+1}+(-1)^q \equiv 0 \mod p </math> | ||
Latest revision as of 17:05, 19 June 2006
I cleaned up the LaTeX a little bit. The LaTeX on the wiki is a little, uh, tough to deal with. For example, it wont allow just p. So instead I had to use {p}. Apparently there are a lot of strings that won't work. If you come across one of them, try working around it with using the curly braces (they don't change how it looks at all). Also, please use \cdot instead of *.
Ok, I will be working on a few examples, and maybe some history to go along.
Solution needed for the first example. --Chess64 17:58, 19 June 2006 (EDT)
The solution to that first example, while maybe clear enough for a Mathlinks forum, seems a little difficult to understand, especially if you're just learning this theorem. How about:
Solution:
since 
so
