Solving Modulo Arithmetic Equations in Z/5Z^2

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Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}
Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??
 
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mtayab1994 said:

Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}



Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??

I get (4, 4) as the only solution. Since -6 ##\equiv -1## (mod 5) ## \equiv ## 4 (mod 5), I think we arrived at the same thing.
 
mtayab1994 said:

Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}
Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??

As Mark already said, your solution is correct. :smile:For your information, \bar or \overline will put a bar on top of a symbol.
Like this: ##\bar{12}## or ##\overline{12}##.Btw, your set should be ##\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}##.

##5\mathbb{Z}## is the set with the number 5 multiplied with any whole number (the 5-folds).

##\mathbb{Z}/5\mathbb{Z}## is the so called quotient set with all 5-folds "divided away".

##A \times B## is the so called cartesian product of sets A and B that consists of ordered pairs.
 
I like Serena said:
As Mark already said, your solution is correct. :smile:


For your information, \bar or \overline will put a bar on top of a symbol.
Like this: ##\bar{12}## or ##\overline{12}##.


Btw, your set should be ##\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}##.

##5\mathbb{Z}## is the set with the number 5 multiplied with any whole number (the 5-folds).

##\mathbb{Z}/5\mathbb{Z}## is the so called quotient set with all 5-folds "divided away".

##A \times B## is the so called cartesian product of sets A and B that consists of ordered pairs.

Thanks for the extra information. Too bad my teacher doesn't tell us any of this stuff.
 
mtayab1994 said:
Thanks for the extra information. Too bad my teacher doesn't tell us any of this stuff.

Ah well, at least we can give some added value then, here on PF. :wink:
 
I like Serena said:
Ah well, at least we can give some added value then, here on PF. :wink:

YEP:redface:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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