, where p and q are distinct primes, it follows

.

Since this isn't true, then no??

- July 10th 2010, 04:49 PMdwsmithx^2\equiv 1 (mod p) and x^2\equiv 1 (mod q), where p and q are distinct primes
, where p and q are distinct primes, it follows

.

Since this isn't true, then no?? - July 10th 2010, 05:04 PMundefined
This is true.

In general and implies . See this MathWorld page, property #11.

I don't see how you made this step. - July 10th 2010, 05:14 PMchiph588@
This is correct and you have the right idea on how to prove it.

See here for full a proof.