Solve a system of equations with Euclid's Algorithm

Question

Find the general solution $(x, y)$ ∈ $\mathbb Z × \mathbb Z$ for the system of equations

$$355x − 113y = 1 \tag 1$$

$$355x + 113y = 1 \tag 2$$

You need one parameter $t ∈ \mathbb Z$ to describe the general solution.

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My attempt:

By Euclid's Algorithm: $\operatorname{gcd}(355,113)=1$

$1 = 355(-7)+113(22)$

So for equation $2, (x,y)=(-7,22)$.

And for equation $1, (x,y)= (-7,-22)$.

So the general solution is $(-7,22t)$, where $t=\{-1,1\}.$

Is my attempt correct? If not, please can you show me how do we solve this?

Answer 1

One solution to $355x + 113y = 1$ is $x = -7$ and $y = 22$. I'm taking your word for that.

So all the other solutions will be $x =-7 +113t$ and $y= 22 - 355t$ where any integer $t$ will yield a solution to $355x + 113y = 1$.

So we need to solve $355(-7+113t) - 113(22-355t) = 1$

That is $-7*355-113*22 + 2*355*113t = 1$ so

$t= \frac {7*355 +113*22+1}{2*355*113}$. Which is impossible as it isn't an integer.

......

Which shouldn't surprise us.

If we hadn't been misled into thinking this was a number theory question to be solve by Euclid's Algorithm we wouldhave noticed if we add the terms together we get

$710x = 2$

And if we subtract the terms we get $226y = 0$.

So $x= \frac 1{355}$ (which if we did our math right is $-7 + 113*\frac {7*355 +113*22+1}{2*355*113}$) and $y =0$ (which if we did our math right is $22 - 355*\frac {7*355 +113*22+1}{2*355*113}$)

Answer 2

Since

$$\color{maroon}{355x + 113y=1}$$

And

$$\color{green}{355x-113y =1}$$

Equating gives us :

$$\color{magenta}{355x +113y = 355x-113y} \implies \color{blue}{113y = -113y} \implies \boxed {\color {red}{y = 0}}$$

Setting $y= 0$ in the above equation,we get

$$ \color{brown}{355x = 1} \implies \boxed{\color{navy}{x = \frac{1}{355}}}$$

Hence there is no simultaneous solution of $x,y \in \mathbb Z.$

You have solved both the equations independently while the solutions are asked for the system of equations.