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I have the following problem: I have to show the following just by using number theory:

$\forall x_o,x_1\in \mathbb{Z}\quad \exists k_0,k_1\in \mathbb{N} \quad \exists y_0,y_1\in \mathbb{Z}: x_0^{k_0}(x_1^{k_1}(1+x_1y_1)+x_0y_0)=0$.

I'm not allowed to use the criteria of coquand and lombardi.

I thought about the following: if $x_0=0$ or $x_1=0$ then you just set $y_0=0$ and the above is shown. So we now set $x_0,x_1\neq 0$. Because $\mathbb{Z}$ is an integrity ring $x_0^{k_0}=0$ or $(x_1^{k_1}(1+x_1y_1)+x_0y_0)=0$. Since $x_0\neq 0$ the other term $(x_1^{k_1}(1+x_1y_1)+x_0y_0)$ has to equal $0$. There are $2$ possible cases:

$1) \ (1+x_1y_1)=0$ and $x_0y_0=0$. This is only possible if $x_1\in\{1,-1\}$. $2)\ x_1^{k_1}(1+x_1y_1)=-x_0y_0$. So my question here is:

Is the following true?:$\ \forall x_0,x_1 \quad \exists k_1,y_1$ so that $x_0\ |\ x_1^{k_1}(1+x_1y_1)$

Can somebody help me please? Thanks in advance.

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  • $\begingroup$ Does $*$ stand for multiplication? $\endgroup$ – Arnaud Mortier May 23 '18 at 21:51
  • $\begingroup$ That's right. I forgot to use cdot $\endgroup$ – Tobi92sr May 23 '18 at 21:58
  • $\begingroup$ In this case you can just use nothing. It's much easier to read. $\endgroup$ – Arnaud Mortier May 23 '18 at 22:00
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Good job for simplifying the problem.

Write $x_0=ab$ where $b$ is the largest factor of $x_0$ that is coprime with $x_1$. In other words, $a$ is the factor of $x_0$ made of all prime numbers that appear in both $x_0$ and $x_1$.

Picking a $k_1$ large enough will make $x_1^{k_1}$ a multiple of $a$. Large enough is, for example, larger than all powers that appear in the prime decomposition of $a$.

Now since $b$ and $x_1$ are coprime, there is a Bezout relation $$px_1+qb=1$$ which you can rewrite as $qb=1-px_1$. Let $y_1=-p$, you now have $$x_0=ab\ |\underbrace{ \ \ x_1^{k_1}\ \ }_{\text{multiple of $a$}}\cdot \underbrace{(1+x_1y_1)}_{\text{multiple of $b$}}$$

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