PROBLEM LINK:
Practice
Contest
Author:Kevin Atienza
Tester:Jingbo Shang and Pushkar Mishra
Editorialist:Kevin Atienza
CREDITS
Special thanks to the author of the problem for providing a detailed explanation.
DIFFICULTY
hard
PROBLEM
Payton numbers are defined, and prime Payton numbers are defined. Given a Payton number, is it prime or not? (Please see the problem statement for more details)
QUICK EXPLANATION
First, the term "prime" defined in the problem statement actually refers to the abstract-algebraic term "irreducible", and the term "prime" itself is defined differently. Thankfully, as we shall see below, the primes and irreducibles in Payton numbers coincide.
Let $\omega$ be a symbol such that $\omega^2 = \omega - 3$. Then there is an isomorphism between the Payton numbers and the Euclidean domain $\mathbb{Z}[\omega]$ given by $\phi(a,b,c) = (33-2a-c)+(b-a)\omega$. This means that the Payton numbers also form a Euclidean domain (thus primes are the same as irreducibles), and that we have reduced the problem to a primality check in $\mathbb{Z}[\omega]$.
In $\mathbb{Z}[\omega]$, all elements $x$ are of the form $x = a + b\omega$ where $a, b \in \mathbb{Z}$. If we define the norm of $x$ (denoted as $Nx$) as $a^2 + ab + 3b^2$, then we can show the following:
- If $x$ is an integer (i.e. $b = 0$), then $x$ is prime if and only if $x$ is a prime integer, and either $x = \pm 2$ or $x \not= \pm 11$ and $-11$ is not a quadratic residue modulo $x$.
- If $x$ is not an integer (i.e. $b \not= 0$), then $x$ is prime if and only if $Nx$ is a prime integer.
This yields an efficient solution to the problem as long as we can quickly check whether an ordinary integer is prime, and we can quickly check whether $-11$ is a quadratic residue modulo a prime. Primality checking is a standard problem which can be solved using the Miller-Rabin test for example, and checking whether $-11$ is a quadratic residue modulo a prime can be done with Euler's criterion.
EXPLANATION
Denote $S$ as the set of Payton numbers.
Primes in $S$
Let $\omega$ be a symbol such that $\omega^2 = \omega - 3$. Then we can define addition and multiplication in the set $\mathbb{Z}[\omega] = \{a + b\omega : a, b \in \mathbb{Z}\}$ naturally, and the resulting structure is actually a principal ideal domain (we'll discuss this below).
In fact, $S$ is isomorphic to $\mathbb{Z}[\omega]$! This means that the structure of $S$ and $\mathbb{Z}[\omega]$ are identical. In other words, $S$ is just $\mathbb{Z}[\omega]$ with its elements simply relabeled or renamed.
Define $\phi$ as a map from $S$ to $\mathbb{Z}[\omega]$ where $\phi(a,b,c) = (33-2a-c)+(b-a)\omega$. Then $\phi$ is an isomorphism, i.e. a bijective map that satisfies:
$$ \phi(a+b) = \phi(a) + \phi(b) $$
$$ \phi(ab) = \phi(a)\phi(b) $$
The inverse of $\phi$ can be written as the following:
$$ \phi^{-1}(a+b\omega) = \begin{cases}
(11-k,11-k+b,11) & \text{if $a$ is even and $a = 2k$} \\\
(4-k,4-k+b,24) & \text{if $a$ is odd and $a = 2k+1$}
\end{cases} $$
This is great, because this means that the properties of $S$ are exactly the same with $\mathbb{Z}[\omega]$. For instance, an element $x$ of $S$ is prime in $S$ if and only if $\phi(x)$ is prime in $\mathbb{Z}[\omega]$. Therefore, we can study the primes in $\mathbb{Z}[\omega]$ instead.
How to find this isomorphism
Now, you might be wondering: how does one discover this isomorphism? To be honest, I don't know of any surefire way to do this. Even inspecting the accepted solutions doesn't help much, because we can't see the process of discovery there. But a possible line of attack might be:
- Play around with multiplication and realize it is commutative, and also associative if you're lucky. This suggests that it might be isomorphic to some well-known ring.
- Realize that multiplication is symmetric in $a$ and $b$, and that the Payton "zero" and "unity" both satisfy $a = b$. This suggests studying the Payton numbers where $a = b$ would be fruitful. If you're lucky enough, you might notice that this subring is isomorphic to the ring of ordinary integers ($(a,a,c)$ pairs with the integer $33-2a-c$).
- Realize that Payton numbers are really just two-dimensional, and (using the knowledge of a subring isomorphic to the integers) guess that they must be isomorphic to a quadratic integer ring. Discovering which ring it is might take some experimentation, but I would guess that the worst is behind at this point.
Primes in $\mathbb{Z}[\omega]$
Notice that one can write $\omega = \frac{1+\sqrt{-11}}{2}$. First, we show that $\mathbb{Z}[\omega]$ is a Euclidean domain. A Euclidean domain $R$ is an integral domain endowed with a Euclidean function, which is a function $f$ from $R$ to $\mathbb{N}$ satisfying the following property:
If $a$ and $b$ are in $R$ and $b \not= 0$, then there exists $q$ and $r$ in $R$ such that $a = bq + r$ and either $r = 0$ or $f(r) < f(b)$.
In the case of $\mathbb{Z}[\omega]$, the Euclidean function we will use is $f(a+b\omega) = a^2+ab+3b^2$. Notice that if one writes $\omega = \frac{1+\sqrt{-11}}{2} = \frac{1}{2}+\frac{\sqrt{11}}{2}i$, then $f(x+yi)$ is just $x^2+y^2$, which is just the square of the distance to the origin $0+0i$ of the complex plane.
Our proof requires a property of the field $\mathbb{Q}[\omega] = \{a + b\omega : a, b \in \mathbb{Q}\}$. Notice that $f$ can be naturally generalized to include $\mathbb{Q}[\omega]$ in its domain, though the codomain of $f$ becomes $\mathbb{Q}$.
Claim
For every element $x$ of $\mathbb{Q}[\omega]$, there exists an element $n$ in $\mathbb{Z}[\omega]$ such that $f(x-n) < 1$.
Proof
Let's call $x$ good if there exists an $n$ in $\mathbb{Z}[\omega]$ such that $f(x-n) < 1$, so we are trying to prove that all elements of $\mathbb{Q}[\omega]$ are good. The following are facts about goodness (simple to show):
- Let $m\in \mathbb{Z}[\omega]$. Then $x\in \mathbb{Q}[\omega]$ is good if and only if $x-m$ is good.
- $x\in \mathbb{Q}[\omega]$ is good if and only if $-x$ is good.
Therefore, if $a+b\omega$ is an element of $\mathbb{Q}[\omega]$, then the element $\lfloor a \rfloor + \lfloor b \rfloor \omega$ is an element of $\mathbb{Z}[\omega]$. So $a+b\omega$ is good if and only if $(a+b\omega) - (\lfloor a \rfloor + \lfloor b \rfloor \omega) = (a-\lfloor a \rfloor) + (b - \lfloor b \rfloor \omega)$ is good. Thus we have reduced the general case to the case where $0 \le a < 1$ and $0 \le b < 1$. Also, note that if $a + b > 1$, then $a+b\omega$ is good if and only if $-a-b\omega$ is good, if and only if $-a-b\omega+(1+\omega) = (1-a)+(1-b)\omega$ is good. Thus we have further reduced the case to the case where $a \ge 0$, $b \ge 0$ and $a + b \le 1$.
Finally, assume $a \ge 0$, $b \ge 0$ and $a + b \le 1$. Then, considering $a+b\omega$ as a point in the complex plane with $\omega = \frac{1+\sqrt{11}i}{2}$, we have that $a + b\omega$ is inside the triangle bounded by vertices $0+0\omega = 0+0i$, $1+0\omega = 1+0i$ and $0+1\omega = \frac{1}{2} + \frac{\sqrt{11}}{2}i$. The vertices are elements of $\mathbb{Z}[\omega]$. Inside this triangle, the point farthest from any vertex is the circumcenter of the triangle, which can be easily calculated as $\frac{1}{2}+\frac{5}{2\sqrt{11}}i$. The norm of this number minus each of the vertices is $9/11 < 1$, so the claim is proven for $a+b\omega$ (take the nearest vertex).
End proof
We can now show that $\mathbb{Z}[\omega]$ is a Euclidean domain. Note that division is defined in $\mathbb{Q}[\omega]$, and that $f(ab) = f(a)f(b)$ for any $a$ and $b$ (Hint: consider $a$ and $b$ as complex numbers).
Claim
$\mathbb{Z}[\omega]$ is a Euclidean domain with the Euclidean function $f(a+b\omega) = a^2+ab+3b^2$.
Proof
Let $a$ and $b$ be in $\mathbb{Z}[\omega]$, and $b \not= 0$. Note that $a/b$ is an element of $\mathbb{Q}[\omega]$. By the claim above, there exists a $q$ in $\mathbb{Z}[\omega]$ such that $f(a/b-q) < 1$. Let $r = a - qb$. Then $f(r/b) < 1$, so
$$ f(r) = f(r/b\cdot b) = f(r/b)f(b) < 1f(b) = f(b) $$
and the Euclidean property is satisfied, QED.
End proof
Being a Euclidean domain is very nice. For instance, greatest common divisors are well-defined and always exist. Also, every Euclidean domain is a principal ideal domain (PID). It is a well-known fact that every PID is a unique factorization domain (UFD), which means that every number can be written uniquely as a product of primes and a unit.
It is also true in UFDs that irreducible elements are the same as prime elements. Note that irreducible and primes are actually distinct, although we are used to the idea that these are the same, because they are the same in the domain of integers. Also, note that the definition given in the problem statement is actually that of irreducible elements, not prime elements, but since these are the same in a UFD, there's no problem.
In number theory, we say that $a$ divides $b$ in $R$, denoted as $a \mid b$, if there is a $c$ in $R$ such that $ac = b$. We say that $x \in R$ is a unit if $x \mid 1$ in $R$ (in the ring of integers, the units are precisely $-1$ and $1$, and in the Payton numbers, $(4,4,24)$ and $(5,5,24)$). A non-unit $p$ is irreducible if $p$ cannot be factorized into non-units (i.e. if $p = ab$, then either $a$ or $b$ is a unit), and a non-unit $p$ is prime in $R$ if $p \mid ab$ implies $p \mid a$ or $p \mid b$.
We now define greatest common divisors. We say that $g$ is a greatest common divisor, or gcd, of $a$ and $b$ if every common divisor of $a$ and $b$ divides $g$. Note that we say a gcd instead of the gcd because there can be many greatest common divisors of $a$ and $b$. For example, if $g$ is a gcd of $a$ and $b$, then $-g$ is also a gcd. However, it is also true that if $g_1$ and $g_2$ are gcd's of $a$ and $b$, then $g_1 = ug_2$ for some unit $u$ (this follows from the easily provable fact that $x \mid y$ and $y \mid x$ implies $x = uy$ for some unit $u$).
We now show that gcd is well-defined:
Claim
Every pair $(a,b)$ of values in a Euclidean domain $R$ has a gcd.
Proof
If $b$ is zero, then $a$ is a gcd, because every element $r$ in $R$ divides $0$ (because $r0 = 0$). Therefore, the common divisors of $a$ and $0$ are simply the divisors of $a$. If $b$ is nonzero, then we prove the claim by induction on $f(b)$ (which makes sense because $\mathbb{N}$ is well-ordered). Let $a = bq + r$ with $r = 0$ or $f(r) < f(b)$.
If $r = 0$, then $b$ is a gcd of $a$ and $b$, because every divisor of $b$ is also a divisor of $bq = a$. If $f(r) < f(b)$, then by induction, $b$ and $r$ has a gcd, say $g$. We claim that $g$ is also a gcd of $a$ and $b$. This is because every common divisor of $a$ and $b$ also divides $a-bq = r$, so by definition of gcd, it also divides $g$, thus proving that $g$ is a gcd of $a$ and $b$. QED.
End proof
Note that this proof is analogous to the Euclidean gcd algorithm. In fact, this is why $R$ is called a Euclidean domain.
We now prove Bezout's identity in Euclidean domains.
Claim
If $g$ is a gcd of $a$ and $b$ in $R$, then there exists $x$ and $y$ in $R$ such that $ax + by = g$.
Proof
If $b = 0$, then $a$ is a common divisor of $a$ and $b$, and since $g$ is a gcd, $a \mid g$, therefore, $g = ac$ for some $c$. This means that $ca + 0b = g$ (i.e. $(x,y) = (c,0)$). If $b$ is nonzero, then we prove the claim by induction on $f(b)$. let $a = bq + r$ with $r = 0$ or $f(r) < f(b)$.
If $r = 0$, then $b$ is a common divisor of $a$ and $b$, and since $g$ is a gcd, $b \mid g$, therefore $g = bc$ for some $c$. This means that $0a + cb = g$ (i.e. $(x,y) = (0,c)$. If $f(r) < f(b)$, then $g$ is also a gcd of $b$ and $r$, so by induction there exists $x'$ and $y'$ such that $x'b + y'r = g$. Then $y'a + (x' - qy')b = g$ (i.e. $(x,y) = (y', x' - qy')$). QED.
End proof
Note that this is analogous to the extended Euclidean gcd algorithm.
Define the conjugate of $a + b\omega$ as $a + b - b\omega$, and denote it as $(a + b\omega)'$. The conjugate has the following nice properties (which are easy to prove and are left as exercises):
- $x'' = x$
- $(x + y)' = x' + y'$
- $(xy)' = x'y'$
- $x$ is prime if and only if $x'$ is prime.
- If $x \mid y$, then $x' \mid y'$
- If $g$ is a gcd of $a$ and $b$, then $g'$ is a gcd of $a'$ and $b'$
- $x$ is an integer if and only if $x' = x$
- $x + x'$ is an integer. This quantity is called the trace of $x$. The trace of $a + b\omega$ is $2a + b$.
- $xx'$ is an integer. This quantity is called the norm of $x$.
The norm is actually important for our purposes. It is denoted as $Nx$ (i.e. $Nx = xx'$), and has the following properties (which are also easy to prove and are left as exercises):
- $N(a+b\omega) = a^2+ab+3b^2$. Note that this coincides with the Euclidean function $f$ we used above.
- $Nx \ge 0$.
- $Nx = 0$ if and only if $x = 0$.
- $Nx = N(x')$
- $x \mid Nx$
- If $x \mid y$, then $Nx \mid Ny$.
- $N(xy) = Nx\cdot Ny$. It follows that if $x$ is a unit, then $Nx = 1$.
- $Nx = 1$ if and only if $x = 1$ or $x = -1$. It follows that the only units in $\mathbb{Z}[\omega]$ are $1$ and $-1$.
We now begin to uncover exactly which elements of $\mathbb{Z}[\omega]$ are prime. We begin with the following:
Claim
If $x$ is a non-unit in $\mathbb{Z}[\omega]$ and $Nx = p$ for some prime integer $p$, then $x$ is prime.
Proof
If $yz = x$, then $Ny\cdot Nz = N(yz) = Nx = p$. Therefore, either $Ny$ or $Nz$ is $1$, so either $y$ or $z$ is a unit. Therefore, $x$ is irreducible, so $x$ is prime, QED.
End proof
Next, we show that the norms of prime elements are actually very limited in form:
Claim
If $x$ is prime in $\mathbb{Z}[\omega]$, then $Nx$ is either a prime integer or a square of a prime integer.
Proof
Factorize the integer $Nx$ as $Nx = p_1\cdot p_2\cdots p_k$. Now $x$ divides $Nx$, and since $x$ is prime, $x$ therefore divides some $p_i$. Therefore, $Nx \mid Np_i = p_i^2$. So $Nx$ can be $1$, $p_i$ or $p_i^2$. But $x$ is not a unit, so $Nx$ is a prime or a square of a prime, QED.
End proof
Now, a normal composite integer is also composite in $\mathbb{Z}[\omega]$, because its integer factorization is also valid in $\mathbb{Z}[\omega]$. However, not all prime integers are primes in $\mathbb{Z}[\omega]$. The following describes a family of prime integers which are not prime in $\mathbb{Z}[\omega]$.
Claim
If $p$ is an odd prime, $p \not= \pm 11$, and if $-11$ has a square root mod $p$, then $p$ is factorable as $p = xx'$, where $x$ and $x'$ are primes in $\mathbb{Z}[\omega]$.
Proof
Let $a$ be a square root of $-11$ mod $p$ (i.e. $a^2 \equiv -11 \pmod{p}$). Note that $p$ divides $a^2+11$.
Now, let $x$ be a gcd of $p$ and $a+1-2\omega$. By Bezout, there exist $A$ and $B$ such that $x = Ap+B(a+1-2\omega)$. By taking conjugates, and noting that $p' = p$ and $(a+1-2\omega)' = (a-1+2\omega)$, we see that $x'$ is a gcd of $p$ and $a-1+2\omega$, and $x' = Ap + B(a-1+2\omega)$.
Thus,
$$ xx' = (Ap+B(a+1-2\omega))(Ap+B(a-1+2\omega)) $$
$$ = A^2p^2+ABp(2a)+B^2(a^2+11) $$
$$ = p[A^2p+AB(2a)+B^2(a^2+11)/p] $$
So $Nx = xx'$ is divisible by $p$, and $x$ is not a unit. Thus $x'$ is also not a unit.
Now, let $g$ be a gcd of $x$ and $a-1+2\omega$. Thus there exists $C$ and $D$ such that $Cx + D(a-1+2\omega) = g$. Now, since $x \mid (a+1-2\omega)$, we have $g \mid (a+1-2\omega)$. Thus, $g \mid (a+1-2w)+(a-1+2w) = 2a$. Thus, $g$ is a common factor of $2a$ and $p$. But $2a$ and $p$ are coprime, so $g \mid 1$, and $gh = 1$ for some $h$.
Thus:
$$ Cx + D(a-1+2\omega) = g $$
$$ Chx + Dh(a-1+2\omega) = gh = 1 $$
$$ Ch(xp) + Dhp(a-1+2\omega) = p $$
Now, $x'$ divides $p$, so $xx'$ divides $xp$. Also, $x$ divides $p$ and $x'$ divides $a-1+2\omega$, so $xx'$ divides $p(a-1+2\omega)$. Therefore, $xx'$ divides $Ch(xp) + Dhp(a-1+2w) = p$.
Now, $Nx = xx'$ divides $p$ and is divisible by $p$, therefore $xx' = p$. Thus, $p$ is product of non-units $x$ and $x'$. Furthermore, $Nx = N(x') = p$, so $x$ and $x'$ are prime, QED.
End proof
The following shows that the converse is also true:
Claim
If $p$ is a prime and $p \not= \pm 11$, and $-11$ doesn't have a square root mod $p$, then $p$ is irreducible in $\mathbb{Z}[\omega]$.
Proof
We prove the contrapositive.
If $p$ is reducible as $xy = p$ with non-units $x$ and $y$, then $Nx\cdot Ny = Np = p^2$. Since $x$ and $y$ are non-units, the only possibility is $Nx = Ny = p$. Now, let $x = a+b\omega$. Then:
$$ p = Nx $$
$$ p = a^2+ab+3b^2 $$
$$ 4p = 4a^2+4ab+12b^2 $$
$$ 4p = (2a+b)^2+11b^2 $$
$$ 0 \equiv (2a+b)^2+11b^2 \pmod{p} $$
$$ (2a+b)^2 \equiv -11b^2 \pmod{p} $$
$$ [(2a+b)b^{-1}]^2 \equiv -11 \pmod{p} $$
This means that $(2a+b)b^{-1} \bmod p$ is a square root of $-11$ mod $p$. Note that $b$ is invertible mod $p$ because if $p \mid b$, then $p$ also divides $p - ab - 3b^2 = a^2$, so $p \mid a$. This means that $p \mid a + b\omega = x$, and $p^2 = Np \mid Nx = p$, a contradiction, QED.
End proof
These two claims cover most of the primes. The only ones not covered are $\pm 2$ and $\pm 11$. Luckily, they can be easily checked by hand: $11$ and $-11$ are not prime because $-11 = (1-2\omega)^2$ and $11 = -(1-2\omega)^2 = (1-2\omega)(1-2\omega)'$, and it can easily be shown that $2$ and $-2$ are prime (Hint: Show that there doesn't exist $x$ such that $Nx = 2$).
Next, we describe the rest of the prime elements of $\mathbb{Z}[\omega]$. Let $x$ be a prime. We have shown above that that $Nx$ should be a prime or a square of a prime. If $Nx$ is prime, we have shown above that $x$ is prime. Now, what if $Nx$ is the square of a prime? The following shows that $x$ is an integer:
Claim
If $x$ is prime and $Nx = p^2$, then $x = \pm p$.
Proof
First, if $p$ cannot be factored in $\mathbb{Z}[\omega]$, then the only nontrivial factorizations of $p^2$ are $p\cdot p$ and $-p\cdot -p$. Since $xx' = p^2$, then $x = \pm p$.
If $p$ can be factorized in $\mathbb{Z}[\omega]$ into two primes $p = yy'$, then $p^2 = yy'yy' = y^2(y')^2$. Thus, the only factors of $p^2$ whose norm is $p^2$ are $\pm y^2$, $\pm yy'$ and $\pm (y')^2$, and $x$ can only be one of those. But none of those are prime, so this case is impossible.
End proof
We can now combine all of the above to characterize the primes in $\mathbb{Z}[\omega]$. Let element $x$ be an element of $\mathbb{Z}[\omega]$. Then:
- If $x$ is not an integer, then $x$ is prime if and only if $Nx$ is a prime integer.
- If $x$ is an integer, then $x$ is prime if and only if $x$ is a prime integer, and either $x = \pm 2$ or $x \not= \pm 11$ and $-11$ is not a quadratic residue modulo $x$.
Checking whether an integer $p$ is prime can be done with Miller-Rabin.
Checking whether $-11$ is a quadratic residue modulo a prime $p$ can be done using the Legendre symbol $\left(\frac{a}{p}\right)$, which is defined as:
$$
\left(\frac{a}{p}\right) = \begin{cases}
0 & \text{if $p \not\mid a$} \\\
1 & \text{if $p\mid a$ and $a$ is a quadratic residue modulo $p$} \\\
-1 & \text{if $p \mid a$ and $a$ is not a quadratic residue modulo $p$}
\end{cases}
$$
By Euler's criterion, this can be calculated in $O(\log p)$ time as:
$$
\left(\frac{a}{p}\right) \equiv a^{\frac{p-1}{2}} \pmod{p}
$$
Note that you could also use the law of quadratic reciprocity to more quickly check whether $-11$ is a quadratic residue: $-11$ is quadratic modulo an odd prime $p$ if and only if $p \equiv 0, 1, 3, 4, 5, \text{ or } 9 \pmod{11}$ (i.e. those odd $p$'s which are quadratic residues modulo $11$).
SUMMARY
We now summarize the algorithm.
def is_prime_integer(k) {
// returns whether k is a prime integer.
// you can use something like Miller-Rabin for this.
}
def is_prime(a,b,c) {
A := 33 - 2*a - c
B := b - a
if (B = 0) {
return |A| = 2 or (
is_prime_integer(|A|) and
modpow(-11,(|A|-1)/2,|A|) = |A|-1
);
} else {
return is_prime_integer(A*A + A*B + 3*B*B);
}
}