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I think that, in this generality, the answer is no.

For instance, take $a=4$, $b \geq 3$. Then $p \geq 5$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.

I think that, in this generality, the answer is no.

For instance, take $a=4$, $b \geq 3$. Then $p \geq 7$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.

I think that, in this generality, the answer is no.

For instance, take $a=4$, $b \geq 2$. Then $p \geq 5$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.

I think that, in this generality, the answer is no.

For instance, take $a=4$, $b \geq 3$. Then $p \geq 5$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.

I think that the answer is no.

For instance, take $a=4$, $b \geq 2$. Then $p \geq 5$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.

I think that, in this generality, the answer is no.

For instance, take $a=4$, $b \geq 2$. Then $p \geq 5$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(\mathbf{Z}_p)^{\times}$ is always a non-square: indeed, if $x=y^2$ then $x^{(p-1)/2}=y^{p-1}=1$, so the order of $x$ divides $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.