Prove that if $X^2 ≡ b\pmod{p^e}$ has a solution for some for $p \geq 3$ , then $X^2 ≡ b\pmod{p^{e+1}}$ has a solution for all $e \geq 1$

Question

Let $p \geq 3$ be a prime and suppose that the congruence $X^2 ≡ b\pmod{p^e}$ (1) has a solution. $\pmod{p^e}$

Prove that for every exponent $e \geq 1$ the congruence $X^2 ≡ b\pmod{p^{e+1}}$ (2) has a solution.

If $\alpha$ satisfies (1), then assume that the solution to (2) has the form $\eta =\alpha+\beta p^e$ for some $\beta$.

Plugging this into (2), you get $\alpha^2 + 2\alpha\beta p^e = b\pmod{p^{e+1}}$.

I have not sure how to get a solution to $\beta$ in terms of $\alpha$. Can someone please go through the remaining steps? It looks like you need to use the fact that: $X^2-b≡ 0\pmod{p^e}$ and $\gcd(p, 2) = 1$.

***This question is taken from Introduction to Mathematical Cryptography question 1.35a.

Use '\pmod{p^e}' to get the parentheses, spacing, and correct typeface. — Arturo Magidin, Nov 15 '22 at 04:51
I think Hensel's lemma is what you are looking for. https://en.wikipedia.org/wiki/Hensel%27s_lemma — cwlo2F, Nov 15 '22 at 04:52
This article explains the Hensel's lemma in elementary number theory: https://brilliant.org/wiki/hensels-lemma/ — cwlo2F, Nov 15 '22 at 04:56
Same as in the linked dupes: use Hensel's Lemma (Newton's method) to lift the root,. — Bill Dubuque, Nov 15 '22 at 07:47

score 0 · Answer 1 · answered Nov 15 '22 at 05:04

Note that the solution is clear if $b \equiv 0 \pmod{p}$, so assume not. In this case, you also have $\alpha \not\equiv 0 \pmod{p}$ from step (1).

From step (1), you get $p^e \mid b - \alpha^2$. Then you can remove $p^e$ from the congruence relation in step (2): $$ 2 \alpha \beta \equiv \frac{b - \alpha^2}{p^e} \pmod{p}. $$ As you mentioned, you have $\gcd(2\alpha, p) = 1$ hence it is invertible on modulo $p$. By multiplying its inverse, you have $\beta$ as result.

Prove that if $X^2 ≡ b\pmod{p^e}$ has a solution for some for $p \geq 3$ , then $X^2 ≡ b\pmod{p^{e+1}}$ has a solution for all $e \geq 1$

1 Answers1