Q:

Solve a system of equations with Euclid's Algorithm

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.

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?

A:

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 $22 - 355*\frac {7*355 +113*22+1}{2*355*113}$)

