Proof that $x^2-98xy+y^2=0, \, x\ne y$ has no soultions over $\mathbb Z$

Question

The problem is to prove that $$x^2-98xy+y^2=0, \, x\ne y$$ has no solutions over integers. I figured out that this equation has a solution over integers if and only if $$x^2-10xy+y^2=0, \, x\ne y$$ has a solution over integers, but don't know what to do next. Could anyone help me with this problem?

Edit: The reduction of the middle term can be made only when we assume an additional condition, namely $x\ne -y$.

Answer 1

$(x, y) \in \Bbb Z^2$ is a solution of $$ \tag{*} x^2 -98xy + y^2 = 0$$ if and only if $u = \frac xy$ is a rational solution of $$ \tag{**} u^2 - 98 u + 1 = 0 \, . $$ The rational root theorem states that the only possible rational roots of $(**)$ are $u= 1$ and $u = -1$, but those are not solutions. Therefore $(*)$ has no solution in the integers.

Alternatively you can mimic the proof of the rational root theorem and show that if $(x, y) \in \Bbb Z^2$ is a solution of $(*)$ with coprime integers $x,y$ then $$ y (98 x -y ) = x^2 $$ implies that $y$ divides $x$. Similarly, $x$ divides $y$, so that $x = \pm y$, which again gives a contradiction.

Answer 2

This is a quadratic in $y$ with solutions $$y=(49\pm20\sqrt{6})x$$ but for $x\in\mathbb{Z}$ we have that $$y=(49\pm20\sqrt{6})x\not\in\mathbb{Z}$$ as $\sqrt{6}$ is irrational. If there was a solution then $$\sqrt{6}=\pm\frac{y-49x}{20x}\in\mathbb{Q}$$ which contradicts the fact that $\sqrt{6}$ is irrational.

Answer 3

Here is my attempt to prove it using divisibility theory.

Let us assume $(x,y)$, where $x \neq y$, is an integer solution of the given equation. First of all, if at least one of $x$ and $y$ is zero, then the other is zero as well. So, we can assume $x,y \neq 0$.

Now, we rewrite the given equation as $(x-y)^2 = 96xy$. Since $3 \mid 96$, we have

$$ \begin{align} & 3 \mid (x-y)^2 \\ \implies & 3 \mid (x-y) \\ \implies & 9 \mid (x-y)^2 \\ \implies & 9 \mid 96xy \\ \implies & 3 \mid xy \\ \end{align} $$

So, $3 \mid x$ or $3 \mid y$. If $3 \mid x$ , then $3 \mid y$ because $3 \mid (x-y)$. Similarly, $3 \mid y$ will lead us to $3 \mid x$. It is now established that $3 \mid x$ and $3 \mid y$.

Let $p,q,k \gt 0$ be largest integers such that $3^p \mid x$ , $3^q \mid y$ , and $3^k \mid (x-y)$. Note that $p$, $q$, and $k$ are finite because none of $x$, $y$, and $x-y$ is zero. Then, $2k$ is the largest integer such that $3^{2k} \mid (x-y)^2$ and $p+q+1$ is the largest integer such that $3^{p+q+1} \mid 96xy$.

We must have that $2k = p+q+1$. Since $2k-1$ is odd, $p$ and $q$ have different parity, and hence can't be equal. WLOG we can assume that $p \lt q$. It is easy to see that $3^p$ is the largest power of three that divides $x-y$. Therefore $k=p$, and we get $p=q+1 \gt q$, which clearly is a contradiction.