Question

Find all integral solutions of x^2 +1≅ 1(mod 5^3).

Answer 1

Answer on question 33901 – Math – Number Theory

Find all integral solutions of $x^2 + 1 \equiv 1 \pmod{5^3}$.

Solution

We have

$$x^2 \equiv 0 \pmod{5^3},$$

Let $f(x) = x^2$

At first consider the equation

$$f(x) = x^2 \equiv 0 \pmod{5}$$

Obviously that the solution of this equation is

$$x \equiv 0 \pmod{5}$$

This is mean that $x = 5t_1, t_1 \in \mathbb{Z}$.

Now consider the equation

$$\frac{f(0)}{5} + f'(0)t_1 = 0 + 0 \equiv 0 \pmod{5}$$

This equation holds for any integer $t_1$. And we get 5 solutions:

$$t_1 \equiv 0 \pmod{5},$$ $$t_1 \equiv 1 \pmod{5},$$ $$t_1 \equiv 2 \pmod{5},$$ $$t_1 \equiv 3 \pmod{5},$$ $$t_1 \equiv 4 \pmod{5}.$$

And for any integer $t_2$ we obtain

$$t_1 = 5t_2, \quad x = 25t_2$$ $$t_1 = 5t_2 + 1, \quad x = 25t_2 + 5$$ $$t_1 = 5t_2 + 2, \quad x = 25t_2 + 10$$ $$t_1 = 5t_2 + 3, \quad x = 25t_2 + 15$$ $$t_1 = 5t_2 + 4, \quad x = 25t_2 + 20$$

Now consider the comparison $f(x) \equiv 0 \pmod{5^3}$ for all these $x$.

1) For $x = 25t_2$ consider the equation

$$\frac{f(0)}{25} + f'(0)t_2 \equiv 0 \pmod{5}$$

It holds for any integer $t_2$.

2) For $x = 25t_2 + 5$ consider the equation

$$\frac{f(5)}{25} + f'(5)t_2 \equiv 0 \pmod{5}$$ $$1 + 10t_2 \equiv 0 \pmod{5}$$ $$10t_2 \equiv -1 \pmod{5}$$

This comparison has no solutions.

3) For $x = 25t_2 + 10$ consider the equation

$$\begin{array}{l} \frac{f(10)}{25} + f'(10)t_2 \equiv 0 \pmod{5} \\ 4 + 20t_2 \equiv 0 \pmod{5} \\ 20t_2 \equiv -4 \pmod{5} \\ \end{array}$$

This comparison has no solutions too.

4) For $x = 25t_2 + 15$ consider the equation

$$\begin{array}{l} \frac{f(15)}{25} + f'(15)t_2 \equiv 0 \pmod{5} \\ 9 + 30t_2 \equiv 0 \pmod{5} \\ 30t_2 \equiv 9 \equiv 4 \pmod{5} \\ \end{array}$$

This comparison has no solutions.

5) For $x = 25t_2 + 20$ consider the equation

$$\begin{array}{l} \frac{f(20)}{25} + f'(20)t_2 \equiv 0 \pmod{5} \\ 16 + 40t_2 \equiv 0 \pmod{5} \\ 40t_2 \equiv -16 \equiv -1 \pmod{5} \\ \end{array}$$

This comparison has no solutions.

Therefor we get 5 solutions for $t_2$:

$$\begin{array}{l} t_2 \equiv 0 \pmod{5}, \\ t_2 \equiv 1 \pmod{5}, \\ t_2 \equiv 2 \pmod{5}, \\ t_2 \equiv 3 \pmod{5}, \\ t_2 \equiv 4 \pmod{5}. \\ \end{array}$$

Only for the 1) case. And we get

$$\begin{array}{l} t_2 = 5t_3, \quad x = 125t_3 \\ t_2 = 5t_3 + 1, \quad x = 125t_3 + 25 \\ t_2 = 5t_3 + 2, \quad x = 125t_3 + 50 \\ t_2 = 5t_3 + 3, \quad x = 125t_3 + 75 \\ t_2 = 5t_3 + 4, \quad x = 125t_3 + 100 \\ \end{array}$$

Answer: $x \equiv 0 \pmod{125}, x \equiv 25 \pmod{125}, x \equiv 50 \pmod{125}, x \equiv 75 \pmod{125}, x \equiv 100 \pmod{125}$.

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