Question

Let $a,b\in R$ . Assume there are positive integers m, n such that role="math" localid="1653715848789" $a^m=b^m,a^n=b^n$and $\left(m,n\right)=1$. Prove that $a=b$. [Remember that negative powers of a and b are not necessarily defined in R , but they do make sense in the field F ; for instance, role="math" localid="1653715929364" $a^{-2}=\frac{1_R}{a^2}$.]

Short answer

It is proved that a = b.

Step-by-step solution

Bezout Identity

Let a and b be integers with the greatest common divisor d . Then, there exists integers or polynomials x and y such that $ax+by=d$.

Proof of a=b

Let $a,b\in R$ and assume the positive integers m and n such that $a^m=b^m,a^n=b^n$and $\left(m,n\right)=1$.

By the Bezout Identity, for $\left(m,n\right)=1$ , there exists an integers x and y such that $mx+ny=1$.

Thus, in the field F , find a as:

$\begin{array}{rcl}a& =& a^{mx+ny}\\ & =& {\left(a^m\right)}^x+{\left(a^n\right)}^y\\ & =& {\left(b^m\right)}^x+{\left(b^n\right)}^y\\ & =& b^{mx+ny}\end{array}$

Since $mx+ny=1$implies that a = b .

Therefore, if $a,b\in R,a^m=b^m,a^n=b^n$and $\left(m,n\right)=1$then,a= b .