Let $R$ be a ring with $n=|R|\geq3$ elements, which has $\frac{n+1}{2}$ squares. Prove that $1+1$ is invertible and $R$ is a field.

I thought that if there are $\frac{n+1}{2}$ squares, then $n+1$ is even, which means $n$ is odd.
Let $k$ be the order of $1$. Obviously, $k|n$, hence $k$ is odd, which implies $(2,n)=1$, resulting that $1+1\neq0,$ i.e. $1+1$ is invertible.

We see that for any $x\in R$, $x\neq-x$, but I got stuck here.

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    $\begingroup$ Is the ring supposed to be commutative? $\endgroup$ – punctured dusk Mar 9 '17 at 21:45
  • $\begingroup$ It's not said, but I guess we can start by proving it is, since every finite field is commutative. $\endgroup$ – ztefelina Mar 9 '17 at 21:49
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    $\begingroup$ That's a good start (+1). We clearly have that $x^2=(-x)^2$. As any $(x,-x)$ pair thus produces only one square, it must be that there are no other repetitions among squares (otherwise there will be too few of them). This implies that $k$ must be a prime. For if $k$ has distinct odd prime factors, then (by the Chinese Remainder Theorem), $1$ is the square of at least four distinct elements. Also, if $k$ is divisible by a square of a prime, then $0$ will be the square of more than one element, again contradicting what we learned. So $k$ is an odd prime. $\endgroup$ – Jyrki Lahtonen Mar 9 '17 at 22:16
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    $\begingroup$ I would argue that you don't know that $n$ is odd because the natural parsing of "if there are ____ squares" would be "at least ___ squares" not "precisely ____ squares.". $\mathbb{Z}$ has $4$ squares. It also has $1000$ squares and $11/2$ squares. $\endgroup$ – Stella Biderman Mar 9 '17 at 22:40
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    $\begingroup$ So $R$ will be an algebra over $\Bbb{F}_p$ without nilpotent elements. There may be a suitable structure theorem for such beasts allowing us to conclude. But, Morpheus beckons here. Stella's interpretation is also interesting. But, if $R$ is a field of $n=4$ elements, it has four squares, $4\ge (4+1)/2$, but $1+1$ is not invertible. IMHO this is a point in favor of the interpretation that the number of squares should be exactly $(n+1)/2$. It may still be true that if the number of squares is at least $(n+1)/2$, then $R$ is a field. But I'm not sure? Anyway, I don't see a proof. $\endgroup$ – Jyrki Lahtonen Mar 9 '17 at 22:46

I am assuming the interpretation that the number of squares is exactly $(n+1)/2$. As the OP observed, this implies that $n$ must be odd, and hence the additive order $k$ of $1$ is also an odd integer.

  1. It follows that $k\cdot x=0$ for all the elements of $R$. If $x\neq0$, this implies that $x\neq -x$ for otherwise the order of $x$ is a factor of both $k$ and $2$, which is absurd.

  2. As $x^2=(-x)^2$ for all $x\neq0$, the $n-1$ non-zero elements of $R$ can have at most $(n-1)/2$ distinct squares. Adding $0^2=0$ to the tally, we can have at most $(n+1)/2$ distinct squares. This means that we must have the following implication:

    If $x^2=y^2$ in the ring $R$, then either $x=y$ or $x=-y$.

  3. It follows that $R$ can have no zero divisors. This is a bit tricky. Assume contrariwise that $uv=0$ for some elements $u,v\in R\setminus\{0\}$. Then
    • We also have $vu=0$. For if $vu\neq0$, then we have $$0^2=0=v0u=vuvu=(vu)^2$$ in violation of the highlighted implication.
    • But if $vu=uv=0$, then we have $$(u-v)^2=u^2-uv-vu+v^2=u^2+v^2$$ as well as $$(u+v)^2=u^2+uv+vu+v^2=u^2+v^2.$$ The highlighted implication then tells us that $u-v=\pm(u+v)$ implying that either $u=0$ or $v=0$. This is a contradiction.
  4. In a ring without zero divisors the cancellation law, $ca=cb\implies a=b$ holds whenever $c\neq0$. Similarly the cancellation law, $ac=bc\implies a=b$ also holds.
  5. Let $r\in R, r\neq0$, be arbitrary. By item 4 the elements $ra$, $a$ ranging over $R$, are all distinct. As $R$ is finite, $1$ is among them, and we can conclude that $r$ has a right inverse. Similarly we see that $r$ has a left inverse. Associativity of the product then implies in the usual way that the left and right inverses must coincide. Therefore $r$ is invertible.
  6. So $R$ is a division ring. By a theorem of Wedderburn every finite division ring is a field, and we are done.
  • $\begingroup$ I wonder whether the piece of information about the number of squares will also let us get away with using Wedderburn. IOW, I hope there might be a lower technology substitute for my step 6. $\endgroup$ – Jyrki Lahtonen Mar 11 at 11:02

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