# mod equations problem...

• Jun 7th 2008, 10:35 AM
aurora
mod equations problem...
Hi!
I need to prove that there is *no* solution to the equation:
x^2 + y^2 = 3(mod4)

I tried to assume there is a solution, meaning, there are X,Y in Z so that:
x^2 + y^2 = 3 + 4*k.

I tried to reach a contradiction, but alas, I did not succeed :-\

Help?
Thank you very much, in advance!
• Jun 7th 2008, 11:26 AM
Moo
Hello,

Quote:

Originally Posted by aurora
Hi!
I need to prove that there is *no* solution to the equation:
x^2 + y^2 = 3(mod4)

I tried to assume there is a solution, meaning, there are X,Y in Z so that:
x^2 + y^2 = 3 + 4*k.

I tried to reach a contradiction, but alas, I did not succeed :-\

Help?
Thank you very much, in advance!

Find all the possibilities.

Attachment 6682

The numbers are the modulus with respect to 4 :)
Does it help ?
• Jun 7th 2008, 07:13 PM
ThePerfectHacker
Quote:

Originally Posted by aurora
Hi!
I need to prove that there is *no* solution to the equation:
x^2 + y^2 = 3(mod4)

I tried to assume there is a solution, meaning, there are X,Y in Z so that:
x^2 + y^2 = 3 + 4*k.

I tried to reach a contradiction, but alas, I did not succeed :-\

Help?
Thank you very much, in advance!

If $x$ is even then $x^2\equiv 0$.
If $x$ is odd then $x^2 \equiv 1$.
Thus, $x^2+y^2 \equiv 0,1,2(\bmod 4)$.
Thus, it is impossible.
• Jun 8th 2008, 01:12 AM
aurora
Thank you very much, both of you!