一道数学题

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一道数学题

Calc number of solution

x^2 - xy - y^2 + yz + z^2 - zx = 0

With

x, y, z \in [1, n], \gcd(x, y, z) = 1 x(x - y) + z(z - x) = y(y - z)

Let

a = x - y, b = y - z, c = z - x

We get

ax + cz = by c = -a - b, y = x - a, z = x - a - b

Thus

ax - (a + b)(x - a - b) = b(x - a) a^2 + 3ab + b^2 = 2bx

If b = 0, x = y = z = 1.

x = \frac{a^2 + 3ab + b^2}{2b}

Because (Considering every possibility can get)

x, y, z \in Odd a, b, c \in Even

Thus

\frac{a^2}{b} \in Even

Let

g = \gcd(a, b), a = ug, b = vg \frac{u^2g}{v} \in \Z

Let

g = kv, a = kuv, b = kv^2 k = \frac{g}{v} = \frac{g^2}{b} \in Even

Thus

x = \frac{k}{2}(u^2 + 3uv + v^2) y = \frac{k}{2}(u^2 + uv + v^2) z = \frac{k}{2}(u^2 + uv - v^2)

Because \gcd(x, y, z) = 1.

\frac{k}{2} = \plusmn 1

Because y > 0, u^2 + uv + v^2 = (u - \frac{1}{2}v)^2 + \frac{3}{4}v^2 \ge 0.

\frac{k}{2} = 1

Thus

x = u^2 + 3uv + v^2 y = u^2 + uv + v^2 z = u^2 + uv - v^2

We have

y = (u - \frac{1}{2}v)^2 + \frac{3}{4}v^2 \frac{3}{4}v^2 \le y, (u - \frac{1}{2}v)^2 \le y |v| \le 2\sqrt\frac{y}{3} \le 2\sqrt\frac{n}{3}, |u - \frac{1}{2}v| \le \sqrt y \le \sqrt n -\frac{2 \sqrt 3}{3}\sqrt n \le v \le \frac{2 \sqrt 3}{3}\sqrt n \frac{1}{2}v -\sqrt n \le u \le \frac{1}{2}v + \sqrt n -\frac{3 + \sqrt 3}{3}\sqrt n \le u \le \frac{3 + \sqrt 3}{3}\sqrt n

In summary

x = u^2 + 3uv + v^2 y = u^2 + uv + v^2 z = u^2 + uv - v^2 1 \le u \le \frac{3 + \sqrt 3}{3}\sqrt n -\frac{2 \sqrt 3}{3}\sqrt n \le v \le \frac{2 \sqrt 3}{3}\sqrt n \gcd(u, v) = 1

Especially when u = 1, v = 0

x = y = z = 1