Show How p Equals 1 or 5 in Z_6 (mod 6)

  • Thread starter Thread starter Pearce_09
  • Start date Start date
Pearce_09
Messages
71
Reaction score
0
Hello,
I am just going to post question i have tried it yet.. only because i don't understand the question.. so if you could help me read the question that would be good.. and if you want to give a hint feel free..

If p does not equal 2, 3 is prime, show that p = 1 or p = 5 in Z_6 (mod 6)

ok what i don't understand is.. is the question saying that p does not also equal 3.. or is it just telling me that 3 is a prime.. cause that obvious..
or is it saying p does not equal 3 but its prime.. I am not sure.. so i can't really go any further.. thank you
 
Physics news on Phys.org
I think it would mean that p is a prime other than 2 or 3.
 
this seems ok (this is not a proof by any means,) but for p=5,7,11,13,17,19,23,29,31 it works. so It seems that our interpetation of the hypothesis on p is correct.
 
awsome..thats what i was thinking to.. thanks!
 
The question just means "if p is a prime not equal to 2 or 3". Exactly what it says. It says nothing about what 2 or 3 are at all, primes or not.

If we were to remove the requirement that p not be 2 or 3 then the statement would read: suppose p is a prime, show p is congruent to 1 or 5 mod 6. And that would be false since there are two primes that are not congruent to 1 or 5 mod 6. However all primes except 2 and 3 are congruent to 1 or 5 mod 6. now let's prove it...

I don't see why this meant you couldn't go further. the question was specifically not about the primes 2 or 3.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top