$x^2+y^2=0$ implies $x^2$ for every $x,y\in R$ is false

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In Exercise 1.9 (iii) in A primer of infinitesimal analysis by John Bell we are asked to show that the following assertion is false: "$x^2+y^2=0$ implies $x^2=0$ for every $x,y\in R$", where $R$ is the fundamental object in any smooth world $\mathbb{S}$, i. e., the smooth real line. Well, for this exercise there is a hint: to use the Principle of Microcancellation, but I just can't even see how to use it to prove that that sentence is false.

Any suggestion, another hint to prove that that sentence is false?

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Assume, for the sake of deducing a contradiction, that for any numbers $x$ and $y$ the implication $x^2 + y^2 = 0 \Rightarrow x^2 = 0$ holds.

Then $\varepsilon \cdot \eta = 0$ for all $\varepsilon,\eta \in \Delta$, by the argument in the following paragraph. However, we know that this statement is wrong (for instance from part (i) of the quoted exercise).

So let $\varepsilon,\eta \in \Delta$ be given. Then $(\varepsilon+\eta)^2 + (\varepsilon-\eta)^2 = 2\varepsilon\eta - 2\varepsilon\eta = 0$. Hence by assumption $(\varepsilon+\eta)^2=2\varepsilon\delta = 0$. Assuming that $2$ in cancellable, we obtain $\varepsilon\delta = 0$.