Use the Euclidean algorithm steps to find ALL the integer solutions of the equation

Question

Use the Euclidean algorithm to find ALL the integer solutions of the equation:

$$5x+72y=1$$

My attempt:

$5x + 72y =1$

$72 = 14 \times 5 + 2 \quad (14~obtained~by~72/5 = 14.4)$

$5 = 2 \times 2 + 1$

Rewrite:

$1 = 5 - 2 \times 2$

$1 = 5 - 2 (72 - 14(5))$

Rewrite:

$5 - 2(72 - 14(5)) = 1$

$5 - 2(72) + 28(5) = 1$

$29(5) - 2(72) = 1$

However, in the solutions this is presented:

$$x = 29 + n \times 72$$

$$y = -2 - n \times 5$$

I have NO idea how they get there.

Answer 1

Suppose you're solving $a x + b y = 1$ in integers. Once you have one solution, say $x = x_0$, $y = y_0$, then $x = x_0 + s$, $y = y_0 + t$ is a solution if and only if $a s + b t = 0$, i.e. $a s = - b t$. Since $a$ and $b$ are coprime (which they must be if $a x + b y = 1$ has a solution), $s$ must be a multiple of $b$ and $t$ must be a multiple of $a$. If $s = k b$, then $t$ must be $-ka$. So the general solution is $$ x = x_0 + k b,\ y = y_0 - k a$$ where $k$ is an arbitrary integer.

Answer 2

It all comes down to $72\cdot 5 - 5\cdot 72=0$ (no duh). And if $5a - 72b =0$ then $a$ must be a multiple of $72$ and $b$ must be a multiple of $5$. (Not so obvious but still pretty easy: $72$ and $5$ are relatively prime so if $5a -72b=0$ we must have $5a$ be a multiple of $72$ and $72b$ be a multiple of $5$, but the relative primeness assures that $a$ is a multiple of $72$ and $b$ is a multiple of $5$.)

So that means if $5x + 72y =1$ then $5(x+72)+72(y-5)= 5x + 72y =5x + 72y +5\cdot 72 - 72\cdot 5 =5x+72y + 0 =5x+72y=1$. And through iteration $5(x + 72n) + 72(y-5n) = 5x + 72y = 1$.

And that's really all there is to it. If $x=29$ and $y=-2$ is one solution. Then $x = 29 +72n$ and $y=-2 -5n$ are an infinite number of solutions.

Slightly trickier, is showing these are the only solutions. But it's not too hard. If $x=29+a$ and $y=-2 - b$ is another solution. The $5(29+a) + 72(-2 -b) = 1 = 5\cdot 29 - 2\cdot 72$ then $5a -72b = 0$ and ... well, we already showed that that means $a$ is a multiple of $72$ and $b$ is a multiple of $5$.

And that's all.

Answer 3

When you have a solution of a diophantine equation $ax+by$, $(x_0,y_0)$ then a general solution is $(x_0+n\frac{b}{d},y_0-n\frac{a}{d})$ ($n$ is any integer) where $d = gcd(a,b)$. Apply this formula to you problem and you get the solution.

Answer 4

You find ONE solution $(x_0,y_0)=(29,-2)$. By the way, I recommend you read it here (after your test :) ).

Let $(x,y)\in \mathbb Z\times \mathbb Z$ another possible solution $$\begin{cases}5x+72y=1 &\color{red}{(R1)}\\ 5x_0+72y_0=1 &\ \color{red}{(R2)}\end{cases}$$

Then $$ 5(x-x_0)+72(y-y_0)=0 \color{red}{(R1-R2)}$$ So, $$\boxed{5(x-x_0)=-72(y-y_0)}\color{blue}{(*)}$$

$\gcd(5,72)=1. $ So $72|x-x_0$ by Gauss's lemma(If an integer $\eta$ divides the product $ab$ of two integers, and is coprime with $a$, then $\eta$ divides $b$) used with $\eta=72, a=5, b=x-x_0, ab=5(x-x_0)$.

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$72|x-x_0$ ie $\exists n\in \mathbb Z:x-x_0=72n$.

Then $y-y_0=-5n$ [Changed $(x-x_0)$ by $72n$ in $\color{blue}{(*)}$]

And finally $$\boxed{\begin{cases}x=29+72n \\ y=-2-5n \end{cases}}, \text{ with }n\in \mathbb Z$$