A domain with $a^{2014}=b^{2014}$ and $a^3=b^3$ Let $(A, +, \cdot) $ be a ring with no zero divisors. If $a, b \in A$ such that $a^{2014}=b^{2014}$ and $a^3=b^3$, then prove that $a=b$.
From the hypothesis we have that $a^{2013}=b^{2013}$
Then I observed that $$a^{2014}-a^{2013}=b^{2014}-b^{2013}\iff a^{2013}(a-1)=b^{2013}(b-1)\iff a=b$$
Is this correct? 
 A: You do not need to work quite so hard, and you need to be a little careful. Note that $2013=671\times 3$ so $$a^{2014} = a^{2013}a = b^{2013}a=b^{2014} $$ 
so that 
$$b^{2013}(a-b)=0$$
Since there are no zero divisors, either $b=0$ whence $a^{2014}=0$ whence $a=0$ and $a=b$; or $a-b=0$ whence $a=b$
I think you do need to treat the $a=b=0$ case separately. Your logic works, but you get $a-1=b-1$ for the case $ab\neq 0$ as an intermediate step. Also you don't need to use $1$ in the proof.
A: You need to point out that a ring having no $0$ divisors always the to "divide both sides"  That if $a*k = b*k$ the either $k = 0$ are $a =b$.  You need to either proof this or state it as a proposition youve proven earlier.  
(Pf: $ak = bk\implies k(a-b) = 0\implies k =0$ or $a-b = 0\implies k=0$ or $a=b$)
Then you can do $a^{2013}(a-1) = b^{2013}(b-1) = a^{2013}(b-1)$ so either $a^{2013} = 0$ or $a-1 = b-1$.  If $a^{2013}=b^{2013}=0$ the $a =b =0$ as there are no zero divisors.
You manipulations could have been simpler.  $a^{2014} = a*a^{2013}$ and $b^{2014} = b*b^{2013} = b*a^{2013}$ so $a*a^{2013} = b*a^{2013}$ so either $a^{2013} = 0$ or $a = b$.
