Prove that a commutative ring F is a field iff each equation ax+b=c
(a,b,c is a member of F and a does not equal 0) has a unique solution in F
well, the condition you have is clearly equaivalent to say that $\displaystyle ax=b$ has a unique solution for any $\displaystyle a \neq 0.$ if $\displaystyle F$ is a field and $\displaystyle 0 \neq a \in F,$ then from $\displaystyle ax=b$ you'll get $\displaystyle x=a^{-1}b$ and you're done.
but the converse is less trivial: so we have that any equation $\displaystyle ax=b, \ a \neq 0,$ has a unique solution in $\displaystyle F.$ first we show that $\displaystyle F$ has no zero divisor: suppose that $\displaystyle uv=0$ and $\displaystyle u \neq 0.$ now the
equation $\displaystyle ux=0$ has two solutions $\displaystyle x=0,v.$ thus $\displaystyle v=0.$ this proves that $\displaystyle F$ has no zero divisor. next we show that $\displaystyle F$ is unitary: choose any $\displaystyle 0 \neq a \in F.$ then $\displaystyle ax_0=a$ for some $\displaystyle x_0 \in F.$ now
let $\displaystyle 0 \neq b \in F.$ then there exists $\displaystyle x_1 \in F$ such that $\displaystyle bx_1=b.$ but then $\displaystyle ab(x_1-x_0)=0.$ since $\displaystyle F$ has no zero divisor, we must have $\displaystyle x_1=x_0.$ clearly $\displaystyle 0x_0=0.$ thus: $\displaystyle \forall b \in F: bx_0=b.$ therefore
$\displaystyle x_0=1_F.$ finally for any $\displaystyle 0 \neq a \in F$ the equation $\displaystyle ax=1$ has a (unique) solution, which means any non-zero element of $\displaystyle F$ is invertible. $\displaystyle \Box$