1. ## Congruence question

Congruences:

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

2. Did you try reading the equation modulo three?

3. Originally Posted by Isomorphism
Did you try reading the equation modulo three?
How would i do that?

4. 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 $\mathbb Z_3$?

5. 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?

6. $0^2=0$
$1^2=1$
$2^2=4=1$

7. 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

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

9. 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??

10. 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)?

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