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I want to show that the diophantine equation does only have the trivial solution $x=y=0$. Since $\text{gcd}(73,-137)=1|0$ this is solveable. So

\begin{align} 137&= 1\cdot73+64\\ 73 &= 1\cdot64+9\\ 64 &= 9\cdot7+1\\ 7 &= 1\cdot7+0 \end{align}

So,

\begin{align} 0&=7-7\cdot1=7-7(64-9\cdot7)=7-7\cdot64+49\cdot9=7-7(137-73)+49(73-64)\\ &=7-56\cdot137+105\cdot73. \end{align}

But this is not in the correct form, the $7$ at the start bothers me and I cant get rid of it without ruining my $137$ or $73$.

How can I fix this?

Parseval
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    $x=137, y=73$ is a nontrivial solution. – lhf Feb 04 '19 at 23:58
  • Seems this site is wrong then. https://planetcalc.com/3303/ – Parseval Feb 05 '19 at 00:00
  • So how exactly is this solved then? If I cant get it in the form where I only have multiples of the coefficients? – Parseval Feb 05 '19 at 00:02
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    That site only seems to find one solution. If there is one solution there are always infinitely many. And if you want $=0$, then $0\cdot 73-0\cdot137=0$ is the most immediate solution and one I should expect such a calculator to find. – Arthur Feb 05 '19 at 00:03

2 Answers2

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The equation can be written $73x=137y$. This means that this number is a common multiple of $x$ and $y$, so it is a multiple of their lowest common multiple, which is $73\cdot137$, because $\gcd(73,137)=1$.

Thus there exists $n$ with $73\cdot137n=73x=137y$. Thus $x=137n$, $y=73n$. Conversely, any pair of numbers of this form is a solution.

When you apply the reverse Euclidean algorithm, you don't start from the line with the zero remainder, but from the line with the least nonzero remainder: \begin{align} 1&=64-9\cdot7\\ &=64-(73-64)\cdot7=73(-7)+64\cdot8\\ &=73(-7)+(137-73)\cdot8\\ &=137\cdot8-73\cdot15 \end{align}

egreg
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  • Ok so now I need to multiply both sides by $0$ thus the trivial solution is $(x^,y^)=(0,0)$. For the general solution, I want to use that for a diophantine equation $ax+by=c$ all soluions are given by $$(x,y)=\left(x^+\frac{b}{gcd(a,b)}n,y^-\frac{a}{gcd(a,b)}n\right),$$ for some integer $n$. Doing this I get that $$(x,y)=(-137n,-73n).$$ Why are my negative, and yours positive? – Parseval Feb 05 '19 at 00:43
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    @Parseval Them being positive or negative is irrelevant, because $n$ can be positive or negative too. – Rhys Hughes Feb 05 '19 at 00:48
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Your method is slightly off. I know what you're attempting, so I'll show the correct method.

Your $gcd=1$. So we take:

$$1=64-(7\cdot 9)$$ $$=64-7(73-64)=8\cdot 64 -7\cdot 73$$ $$=8(137-73)-7\cdot 73=8\cdot137-15\cdot73$$

There's your first solution.

The rest are given by:

$$(8+73k)\cdot 137 -(15+137k)\cdot 73$$ as this ensures equality is kept.

Rhys Hughes
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  • I don't see how this is. I think you forget that the equation is equal to $0$ and not $1$. – Parseval Feb 05 '19 at 00:28
  • Yes. I misread the question, this is the method you were trying to use, but the solutions to your equation are clearly $x=137k, y=73k$ This method is what you use when you are solving $ax+by=gcd(a,b)$ – Rhys Hughes Feb 05 '19 at 00:41
  • Well, I wanted to know how to use that method to show that the solutions are those you stated. – Parseval Feb 05 '19 at 00:45
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    You have $ax-by=0\to ax=by \to x=\frac ba\cdot y$. It suffices that if $y=ak$ then $x=bk$ follows for any integer $k$. – Rhys Hughes Feb 05 '19 at 00:47