Congruences:

Show by congruences that

q^2 + 3r^2 + 6s^3 – 9p^5 = 2

has no solutions in Z.

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- Jan 5th 2009, 11:32 PMmaths900Congruence 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 AMIsomorphism
Did you try reading the equation modulo three?

- Jan 6th 2009, 03:45 AMmaths900
- Jan 6th 2009, 04:18 AMvincisonfire
In $\displaystyle \mathbb Z_3 $ the equation becomes $\displaystyle q^2 =2 $ because 3,6 and 9 are in the equivalence class of 0.

What are the squares in $\displaystyle \mathbb Z_3 $? - Jan 6th 2009, 06:44 AMmaths900
- Jan 6th 2009, 07:06 AMvincisonfire
$\displaystyle 0^2=0$

$\displaystyle 1^2=1$

$\displaystyle 2^2=4=1$ - Jan 6th 2009, 07:29 AMmaths900
- Jan 6th 2009, 07:32 AMvincisonfire
No problem. We all have to start somewhere. It will seem simple to you in few weeks from now.

- Jan 6th 2009, 07:34 AMmaths900
- Jan 6th 2009, 08:01 AMmaths900
- Jan 6th 2009, 08:46 AMvincisonfire
Yes, because there is no elements in mod3 such that its square is 2.