Congruence question

**maths900** · January 5th 2009, 11:32 PM

Congruences:

Show by congruences that
q^2 + 3r^2 + 6s^3 – 9p^5 = 2
has no solutions in Z.

**Isomorphism** · January 6th 2009, 12:11 AM

Did you try reading the equation modulo three?

**maths900** · January 6th 2009, 03:45 AM

Originally Posted by Isomorphism

Did you try reading the equation modulo three?

How would i do that?

**vincisonfire** · January 6th 2009, 04:18 AM

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 $\mathbb Z_3$ ?

**maths900** · January 6th 2009, 06:44 AM

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?

**vincisonfire** · January 6th 2009, 07:06 AM

$0^2=0$
$1^2=1$
$2^2=4=1$

**maths900** · January 6th 2009, 07:29 AM

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

**vincisonfire** · January 6th 2009, 07:32 AM

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

**maths900** · January 6th 2009, 07:34 AM

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??

**maths900** · January 6th 2009, 08:01 AM

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)?

**vincisonfire** · January 6th 2009, 08:46 AM

Yes, because there is no elements in mod3 such that its square is 2.

Thread: Congruence question