Prove that if $a$ and $b$ are distinct elements of a group $G$, then $a^2 \ne b^2$ or $a^3 \ne b^3$. Prove that if $a$ and $b$ are distinct elements of a group $G$, then $a^2 \ne b^2$ or $a^3 \ne b^3$.
Im not really sure how to approach this, any thoughts?
Thanks
 A: the point is that $2,3$ are coprime so every integer can be expressed in the form $2x + 3y$ by Bezout.
Let $a^2=b^2$ and $a^3=b^3$ then for any $n$ we can pick $x,y$ such that $2x+3y = n$ and then $a^{2x+3y}=b^{2x+3y}$ since $(a^{2})^x(a^{3})^y = (b^{2})^x(b^{3})^y$
A: Contrapositively, it is $\rm\: a^2 = b^2,\ a^3 = b^3\:\Rightarrow\: a = b,\:$ with easy proof:
$\qquad\qquad\qquad\qquad\quad\begin{eqnarray}\rm a^2 = b^2\\ \rm \color{#C00}{a^3 = b^3}\end{eqnarray}$ $\ \Rightarrow$ $\begin{eqnarray} &\rm a^{-2}& = &\rm b^{-2}& \\
\rm a = \color{#C00}{a^3}\!\! &\rm a^{-2}&\rm  = \color{#C00}{b^3}\! &\rm b^{-2}&\rm = b\ \ \ \ \ {\bf QED} \end{eqnarray}$
Alternatively, more conceptually, the set $\rm\,S\,$ of integers $\rm\,n\,$ with $\rm\:a^n = b^n\:$ is nonempty ($\rm\,2,3\in S)$ and, furthermore, $\rm\: S\: $ is $ $ closed under subtraction $ $  since 
$$\rm j, k \in S\ \Rightarrow\ a^j = b^k,\,\ a^{k} = b^{k}\ \Rightarrow\ a^{-k} = b^{-k}\ \Rightarrow\ a^{j-k} = b^{j-k} \Rightarrow\ j-k\in S $$
Hence, by the (subtractive) Euclid algorithm, $\rm\,S\,$ closed under subtraction $\Rightarrow$ $\rm\, S\,$ closed under gcd. Thus since  $\rm\,S\,$ contains two coprime integers, $\rm\,S\,$ contains their gcd $ = 1,\:$ thus $\rm\:\color{#C00}1\in S\:\Rightarrow\: a^{\color{#C00}1} = b^{\color{#C00}1}.$
Remark $\ $ If you know about groups or ideals, then the above may be viewed as noticing that the set $\rm\,S\,$ of equalizing exponents, being closed under subtraction, is an additive subgroup of $\:\Bbb Z,\:$ therefore is cyclic, generated by the gcd of its elements. Such cyclic groups (and ideals) are hidden everywhere in elementary number theory. Great insight is gained by bringing this hidden structure to the fore, esp. explicitly recognizing certain ubiquitous forms of these structures, for example denominator ideals and  order ideals (annihilators).
A: What would be happen if $$a^2=b^2,~ \wedge~~ a^3=b^3$$ simultaneously? It would happen $a=b$. A nice contradiction!
A: HINT: If $a^2=b^2$ and $a^3=b^3$, then $b^2=a^2=a^3a^{-1}=b^3a^{-1}$. What does this tell you about $ba^{-1}$?
