If mapping is a homomorphism, then cardinality of ring is uniquely determined

Question

Let $(R,+,\cdot)$ be a ring with identity where $0\neq 1$ and $|R|>2$. Define $\varphi :R\to R$ with $\varphi (x) := x^{10}$.

Show that if $\varphi$ is a group endomorphism (of $(R,+)$), then $R$ consists of precisely four elements.

Ignore this topic

Answer 1

This is false. For instance, if $R$ is a product of any number of copies of $\mathbb{Z}/2$, then $x^2=x$ for all $x\in R$ and so $\varphi$ is the identity map, which is a group endomorphism.