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This question is the true version I wanted to ask of this question.

Say $p$ is an odd prime number, $k$ a positive integer and $p^{\tau} || k$. Let $\gamma = \tau + 1$. I would like to prove

If $y \in \mathbf{Z}$ is a $k$-th power modulo $p^\gamma$, then it is also a $k$-th power modulo $p^t$ for any $t \geqslant \gamma$.

I don't know how standard this fact is. Stated like this, it seems to be a rather simple fact. However, I do not find any simple proof for this fact, and I tried to write a messy proof (see post quoted above) but nothing that convinces me or shed lights on what is happening.


This questions comes from my misunderstanding of the following statement, from Vaughan:

Vaughan's statement, Circle method book

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  • $\begingroup$ You need $y$ coprime to $p$ for the highlighted statement to be true. $\endgroup$ – Bart Michels Feb 14 at 9:08
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Hensel's lemma for singular points implies that:

When $y$ is coprime to $p$ and a $k$th power residue mod $p^\gamma$, with $\gamma > 2 \tau$, then $y$ is a $k$th power residue mod all $p^t$.

The philosophy behind Hensel's lemma is to use Newton's method, which requires that an approximation to a root is sufficiently close to a root relative to the derivative. Here: $f(x) = x^k-y$, $|f'(x)| = p^{-\tau}$, and one needs $|f(x)| < |f'(x)|^2$, i.e. $\gamma > 2\tau$.

In this particular case, the proof of Hensel's lemma can be adapted to show that $|f(x)| < |f'(x)|$ is sufficient, i.e. that $\gamma > \tau$ is sufficient.

Here we go. Take $\gamma > \tau$ and $x \in \mathbb Z$ with $x^k -y \equiv 0 \pmod {p^\gamma}$. We will find $a \in \mathbb Z$ with $(x+ a p^{\gamma - \tau})^k - y \equiv 0 \pmod{p^{\gamma+1}}$. By induction, it then follows that $y$ is a $k$th power modulo all $p^t$. Divide by $p^\gamma$ to obtain the condition $$\frac{x^k - y}{p^\gamma} + kx^{k-1}ap^{-\tau} + \sum_{j = 2}^k \binom k j x^{k-j} a^j p^{j(\gamma-\tau)-\gamma} \equiv 0 \pmod p \,.$$ The coefficient of $a$ in the second term is invertible mod $p$ (this is why we took the power $p^{\gamma-\tau}$ in the beginning of the proof). Let's show that all terms in the sum are divisible by $p$. We can then take $$a = - \frac{x^k - y}{p^\gamma} \cdot (kp^{-\tau}x^{k-1})^{-1} \bmod p \,.$$ Write the $j$th term as $$\frac{k}j \binom{k-1}{j-1} x^{k-j} a^j p^{j(\gamma-\tau)-\gamma} = \frac{kp^{-\tau}}j \binom{k-1}{j-1} x^{k-j} a^j p^{(j-1)(\gamma-\tau)} \,.$$ We want to show that $v_p(j) < (j-1)(\gamma-\tau)$. Because $\gamma > \tau$, it suffices to show that $v_p(j) < j-1$. Because $j < p^{j-1}$ (here we use that $p >2$ for the only time), $j$ cannot be divisible by $p^{j-1}$.

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  • $\begingroup$ Thank you very much for your answer. However, it is not sufficient to prove Vaughan's statement, right? (he only requires $\gamma = \tau + 1$) $\endgroup$ – Gory Mar 18 at 15:39
  • $\begingroup$ @Gory the answer addresses this $\endgroup$ – hunter Mar 18 at 15:43
  • $\begingroup$ @Gory My answer is a proof of that statement. You can skip the first three paragraphs and start reading from "Here we go", which is where the proof begins. $\endgroup$ – Bart Michels Mar 18 at 15:56

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