Hello, Codeforces!
Today I'm going to talk about an unpopular technique in number theory.
Definition and elementary properties
Def. Let $$$p > 2$$$ be a prime number, then we will call all $$$1 \le x \le p - 1$$$ modulo $$$p$$$ such that the equation $$$a^2 \equiv x \pmod{p}$$$ has solutions and quadratic non-subtraction otherwise. Note that $$$0$$$ is neither quadratic deduction nor quadratic non-deduction.
Theorem: Quadratic deduction and cradratic non-deduction are equally divided.
It turns out that all numbers split into pairs $$$x$$$ and $$$-x$$$, so each quadratic deduction has exactly $$$2$$$ roots, with each number being the root of some quadratic deduction $$$\implies$$$ of quadratic deductions of exactly $$$\frac{p - 1}{2}$$$ $$$\implies$$$ of quadratic non-deductions $$$p - 1 - \frac{p - 1}{2} = \frac{p - 1}{2}$$$.
Theorem: Denote by
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.
Let $$$x$$$ and $$$y$$$ be quadratic deductions, let $$$a^2 \equiv x \pmod{p}$$$
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$$$. Then $$$(ab)^2 \equiv xy \pmod{p}$.$$$H \cdot B = H$$$
Let $$$x$$$ be a quadratic nondeduction and $$$y$$$ be a quadratic deduction, we prove that $$$\frac{1}{y}$$$ is also a quadratic deduction. Let $$$b^2 \equiv y \pmod{p}$$$, then $$$(\frac{1}{b})^2 \equiv \frac{1}{y} \pmod{p}$$$.
$$$x \cdot y \equiv z \pmod{p}$$$
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is a quadratic deduction, then $$$x \equiv z \cdot \frac{1}{y} \pmod{p}$$$ $$$\implies$$$ $$$x$$$ is quadratically deductible, since $$$z$$$ is quadratically deductible and $$$\frac{1}{y}$$$ is quadratically deductible. The contradiction $$$\implies$$$ $$$z$$$ is quadratically non-deductive.$$$H \cdot H = B$$$
Let $$$x$$$ and $$$y$$$ be quadratic nondeductions, $$$x \cdot y \equiv z \pmod{p}$$$ and $$$z$$$ a quadratic nondeduction, and $$$a_1, a_2, a_3 \cdots, a_{\frac{p - 1}{2}}$$$
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. All numbers in this set are quadratic nonconvex, but there are $$$2$$$ equal numbers in the set $$$\frac{p + 1}{2}$$$, and this is more than the quadratic nonconvex $$$\implies$$$ in the set. Note that if $$$xk \equiv xl \pmod{p}$$$ and $$$x$$$ is not a multiple of $$$p$$$, then $$$k \equiv l \pmod{p}$$$. So in the set $$$a_1, a_2, a_3 \cdots, a_{\frac{p - 1}{2}}, y$$$ there are equal numbers, but in this set all numbers are different ($$$a_i \ne a_j$$$ since we took different quadratic deductions, and $$$y \ne a_i$$$ since $$$y$$$ is quadratic non-deduction and $$$a_i$$$ is quadratic deduction). Contradiction, so $$$z$$$ is a quadratic deduction. .With these two theorems, we can already solve 103428K - Tiny Stars.
How to check if the number is deductive or non-deductive?
There are several ways to check if a number is quadratic deduction. In this blog, we will look at just one of them, you can read about [Gauss's lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(number_theory)) and law of quadratic reciprocity.
Euler's criterion
$$$a$$$ is a quadratic deduction modulo $$$p$$$ if and only if $$$a^{{\frac{p - 1}{2}}} \equiv 1 \pmod{p}$$$, and a quadratic non-deduction if and only if $$$a^{\frac{p - 1}{2}} \equiv -1 \pmod{p}$$$.
Let $$$a$$$ be a quadratic deduction and $$$x^2 \equiv a \pmod{p}$$$, then $$$a^{\frac{p - 1}{2}} \equiv 1 \pmod{p}$$$, since $$$(x^2)^{\frac{p - 1}{2}} = x^{p - 1} \equiv 1 \pmod{p}$$$, and this is true by Fermat's small theorem.
Consider the polynomial $$$x^{{\frac{p - 1}{2}} - 1$$$ modulo $$$p$$$, its degree is $$$\frac{p - 1}{2}$$$ and we have already found it has $$$\frac{p - 1}{2}$$$ roots $$$\implies$$$ no more roots $$$\implies$$$ if $$$b$$$ is quadratically non-degenerate, then $$$b^{\frac{p - 1}{2}} \equiv -1 \pmod{p}$$$.
Consequence: $$$-1$$$ is quadratically deductible if and only if $$$p = 4k + 1$$$, for some natural $$$k$$$, and quadratically non-deductible if and only if $$$p = 4k + 3$$$, for some natural $$$k$$$.
Clearly, the complexity of the number check is $$$O(\log_2p)$$$.
bool check(int a, int p) { return power(a, (p — 1) / 2, p) == 1; } ~~~~
Finding $$$i$$$ modulo $$$p$$$
Def. $$$i$$$ is such a number that $$$i^2 = -1$$$ $$$\implies$$$ $$$i$$$ modulo $$$p$$$ let us call such a number that $$$i ^ 2 \equiv -1 \pmod{p}$$$.
Algorithm: If $$$p = 4k + 3$$$ for some natural $$$k$$$, then there is no such $$$i$$$. If $$$p = 2$$$, then $$$i = 1$$$. Otherwise consider the quadratic non-deduction of $$$a$$$, by Euler's criterion we know that $$$a^{\frac{p - 1}{2}}} \equiv -1 \pmod{p}$$$, then $$$a^{\frac{p - 1}{4}} \equiv i \pmod{p}$$$. All that remains is to find a quadratic non-deduction, for this we take a random $$$1 \le a \le p - 1$$$ and check it for $$$O(\log_2p)$$$, if it is a quadratic deduction then take another random $$$a$$$. Since the deductions and non-deductions are equal, we need $$$O(1)$$$ of such checks, so the total complexity of the algorithm is $$$O(\log_2p)$$$.
int power(int a, int x, int mod) { int res = 1; while (x) { if (x % 2 == 1) { res = 1ll * res * a % mod; } a = 1ll * a * a * a % mod; x /= 2; } return res; }
bool check(int a, int p) { return power(a, (p — 1) / 2, p) == 1; }
int find_i(int p) { if (p % 4 == 3) { return -1; } if (p == 2) { return 1; } while (true) { int a = mt() % (p — 1) + 1; if (!check(a, p)) { return power(a, (p — 1) / 4, p); } } } ~~~~
