# Congruence question

• Jan 5th 2009, 11:32 PM
maths900
Congruence question
Congruences:

Show by congruences that
q^2 + 3r^2 + 6s^3 – 9p^5 = 2
has no solutions in Z.
• Jan 6th 2009, 12:11 AM
Isomorphism
Did you try reading the equation modulo three?
• Jan 6th 2009, 03:45 AM
maths900
Quote:

Originally Posted by Isomorphism
Did you try reading the equation modulo three?

How would i do that? (Thinking)
• Jan 6th 2009, 04:18 AM
vincisonfire
In $\mathbb Z_3$ the equation becomes $q^2 =2$ because 3,6 and 9 are in the equivalence class of 0.
What are the squares in $\mathbb Z_3$?
• Jan 6th 2009, 06:44 AM
maths900
Quote:

Originally Posted by vincisonfire
In $\mathbb Z_3$ the equation becomes $q^2 =2$ because 3,6 and 9 are in the equivalence class of 0.
What are the squares in $\mathbb Z_3$?

Would it be plus or minus root 2? :confused:
• Jan 6th 2009, 07:06 AM
vincisonfire
$0^2=0$
$1^2=1$
$2^2=4=1$
• Jan 6th 2009, 07:29 AM
maths900
Quote:

Originally Posted by vincisonfire
$0^2=0$
$1^2=1$
$2^2=4=1$

and these are not solutions in the set Z? sorry im new to number theory
• Jan 6th 2009, 07:32 AM
vincisonfire
No problem. We all have to start somewhere. It will seem simple to you in few weeks from now.
• Jan 6th 2009, 07:34 AM
maths900
Quote:

Originally Posted by vincisonfire
No problem. We all have to start somewhere. It will seem simple to you in few weeks from now.

I hope so.

Would i not be able to show it has no solutions in Z without using congruences??
• Jan 6th 2009, 08:01 AM
maths900
Quote:

Originally Posted by vincisonfire
In $\mathbb Z_3$ the equation becomes $q^2 =2$ because 3,6 and 9 are in the equivalence class of 0.
What are the squares in $\mathbb Z_3$?

Here when you say q^2=2, do you mean q^2 is congruent to 2(mod3)?
• Jan 6th 2009, 08:46 AM
vincisonfire
Yes, because there is no elements in mod3 such that its square is 2.