Exercise 4.24

Proof. If $a\wedge b\nmid c$, the two equations have no solution. So we can suppose $a\wedge b\mid c$, and after division by $\delta =a\wedge b$, we obtain an equation $a^{\prime }{x}^m+b^{\prime }{y}^n=c^{\prime }$, $a^{\prime }=a∕\delta ,b^{\prime }=\mathit{b\delta },c^{\prime }=\mathit{c\delta }$, and $a^{\prime }\wedge b^{\prime }=1$. So it remains to prove that $a{x}^m+b{y}^n\equiv c(\mathit{mod}p)$ has the same number of solutions as $a{x}^{m^{\prime }}+b{y}^{n^{\prime }}\equiv c(\mathit{mod}p)$ when $a\wedge b=1$.

In this case the equation $\mathit{au}+\mathit{bv}=c$ has solutions. Let $N$ be the number of solutions $(\bar{x},\bar{y})$ of the equation $\bar{a}{\bar{x}}^m+\bar{b}{\bar{y}}^n=\bar{c}$, and $N^{\prime }$ be the number of solutions $(\bar{x},\bar{y})$ of the equation $\bar{a}{\bar{x}}^{m^{\prime }}+\bar{b}{\bar{y}}^{n^{\prime }}=\bar{c}$. Then

The same is true for $N^{\prime }$, so it is sufficient to prove that

where $m^{\prime }=m\wedge (p-1)$, and a similar equality for the equation ${\bar{y}}^n=\bar{v}$.

Let $\bar{g}$ be a generator of ${𝔽}_p^{\text{∗}}$. Write $\bar{u}={\bar{g}}^r,r\in \mathbb{N}$.

Therefore

and similarly

Since $m^{\prime }\wedge (p-1)=(m\wedge (p-1))\wedge (p-1)=m\wedge (p-1)$, these two conditions are equivalent, so these two sets are empty for the same values of $\bar{u}$.

Let $\bar{u}$ be such that $\{\bar{x}\in {𝔽}_p|{\bar{x}}^m=\bar{u}\}\ne \varnothing$, and $x_0$ be a fixed solution of ${\bar{x}}^m=\bar{u}$.

Write $\bar{x}={\bar{g}}^k,\overline{x_0}=g^{k_0}$. Let $d=m\wedge (p-1)(=m^{\prime })$.

As $g$ is a primitive root modulo $p$, the distinct solutions are $x_0,x_0{g}^{\frac{p-1}{d}},\dots ,x_0{g}^{k\frac{p-1}{d}},\dots x_0{g}^{(d-1)\frac{p-1}{d}}$, therefore in this case

As $m^{\prime }\wedge (p-1)=m\wedge (p-1)$,

So $N=N^{\prime }$ : $a{x}^m+b{y}^n\equiv c(\mathit{mod}p)$ has the same number of solutions as $a{x}^{m^{\prime }}+b{y}^{n^{\prime }}\equiv c(\mathit{mod}p)$, where $m^{\prime }=(m,p-1)$ and $n^{\prime }=(n,p-1)$. □