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In general how to solve this equation:(with Hensel's lemma and without it) $2x^3+7x-4\equiv0 \pmod{25} $

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    The crude way without Hensel is to note that $1$ is the only solution mod $5$, and then try $1$, $6$, $11$, $16$, $21$ modulo $25$.2012-12-07

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Without Hensel's lemma: First solve $2x^3+7x-4\equiv 0\pmod 5$, getting a solution $x_0$.

Then write $x=x_0+5k$ and substitute into the equation:

$2(x_0+5k)^3+7(x_0+5k)-4\equiv 0\pmod {25}$

solving for $k$. This turns out to be linear in $k$ because the high powers of $k$ have coefficients that are divisible by $25$.

With Hensel's lemma: Hensel's lemma is a shorter version of this, and can be written much like Newton's method:

$p(x)=2x^3+7x-4$

Let $x_0$ be a solution of $p(x)\equiv 0 \pmod 5$.

Then $x_1=x_0 + \frac{p(x_0)}{p'(x_0)}$

Where by the fraction, we really mean multiplication by an inverse of $p'(x_0)$ modulo $5$. Clearly, that's only going to be possible of $p'(x_0)\not\equiv 0\pmod 5$.

Then $x_1$ is a solution $\pmod {25}$.

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    That was what I got. There is only one solution $\pmod 5$, and only one $k$, so yes, only one solution $\pmod {25}$.2012-12-07