1

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?

Javier Alvez
  • 15
  • 1
    Simpler: cancel $, \large a^{2013}=b^{2013}$ from $,\large a^{2014} = b^{2014}$ to get $,\large a = b\ $ [$after$ handling $,a = 0\ (!\iff b= 0),$]. More generally it is true for coprime powers - see here. – Bill Dubuque Aug 04 '19 at 20:33

2 Answers2

1

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.

Mark Bennet
  • 100,194
0

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 $a = b$.

fleablood
  • 124,253