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Proof help watch
- Thread Starter
- 09-02-2010 19:58
- 09-02-2010 20:14
The second case you need to consider is n|a and n|c. Because a and b are symmetric, you can consider the other, third, case (namely, n|b, n|c) as following from this second case.
Bundling up the other two cases by just considering the second case is what is known as a "without loss of generality" (WLOG) step.
- 09-02-2010 20:19
Isn't it similar to replacing a and b with np and nq, but instead of getting c^2 = (n^2)(p^2 + q^2), you get a^2 = (n^2)(p^2 - q^2)?