Spanning set

Let $S = \{0,1,2,\dots,10\}$.

A set of numbers $A \subset S$ is called "spanning" if the set $\{ x, x+1, x+2, x+3 \mod 11 | x \in A \} = S$.

And a set of numbers $B \subset S$ is called "potentially spanning" if there is an integer $k$ such that the set $\{kb \mod 11 | b \in B \}$ is spanning.

Prove that every subset of $S$ with four elements are potentially spanning, and prove that there exists a subset of $S$ with three elements that is not potentially spanning.

Solution First, we show that there is a subset of three elements that is not potentially spanning.

Note that for a set to be spanning, it has to be $\{c,c+4, c+8\}$ for some $c$ (here we assume everything is modulo 11, so we don't write it for notational simplicity). This is because the maximum distance between two adjacent element is 4, so the distances between 3 elements must be 4+4+3 (or its rotation).

Now we claim that a set $\{x,y,z\}$ is potentially spanning if and only if there exists $r \neq 0$ and $s$ such that $\{x,y,z\} = \{r+s, 2r+s, 3r+s \mod 11\}$. First, clearly if a set is spanning, then all of its "rotations" are also spanning sets. Likewise, if a set is potentially spanning, then all of its rotations are also potentially spanning. So wlog we may assume that $s=0$. Then the set $\{r,2r,3r\}$ is potentially spanning because we can always take $k = 4r^{-1} \mod 11$ so that $\{kr,2kr,3kr\} = \{4,8,12=1\}$ a spanning set. The inverse modulo is guaranteed to exist because 11 is prime and $r \neq 0$.

Now the other direction. Suppose $\{x,y,z\}$ is potentially spanning, such that $\{kx,ky,kz\}$ is spanning. As we've established above, it must have form $\{c,c+4, c+8\}$ for some $c$. Thus we can take $r = 4k^{-1} \mod 11$ and $s = (c-4)k^{-1} \mod 11$ so that $\{k(r+s),2kr,3kr\} = \{4+ks,8+ks,12+ks\} = \{c,c+4,c+8\}$.

So we have shown that the potentially spanning sets are generated by (and only by) the number of $(r,s)$ pairs we can choose. There are 10 choices for $r$ and 11 for $s$, so there are at most 110 potentially spanning sets. We say at most because some choices of $r$ and $s$ may "collide" (result in the same spanning triplet). However, there are ${11 \choose 3} = 165$ possible triplets, so there are some triplets that are not generated by the $(r,s)$ construction above and thus not potentially spanning.

Now we show that a subset $B = \{b_1,b_2,b_3,b_4\}$ is always potentially spanning. Note that if any three of $b_i$s form a potentially spanning triplet then $B$ is potentially spanning. In particular, if $B$ contains an arithmetic sequence modulo 11 then $B$ is potentially spanning. So we assume that $B$ is free from arithmetic progression.

Now let $d_i = b_{i+1} - b_i$ (of course we mean $b_5 = b_1$. Then among all four $d_i$s, let $d$ be the maximal distance. If the maximal distance is 4, then we are done because $B$ itself is spanning. So we assume that the maximal distance is at least 5. Wlog, we may assume that this maximal distance is $d_4$ and that $b_1 = 0$. So this means $0=b_1 < b_2 < b_3 < b_4 = 11-d$ and $b_i$ must not contain arithmetic sequence. It's easy to see that the only possible combinations are:

Case 1: $d=5$, so $0=b_1 < b_2 < b_3 < b_4 = 6$. The only combinations are $\{0,1,4,6\}, \{0,1,5,6\},\{0,2,5,6\}$. All the other combinations contain an arithmetic sequence. But in the first two cases, we can take $k=8$ to get: $\{0,4,8,10\}, \{0,4,7,8\}$ which contains arithmetic sequence thus potentially spanning, and for the third case we can take $k=3$ to get $\{0,4,6,7\}$ which is spanning.

Case 2: $d=6$, so $0=b_1 < b_2 < b_3 < b_4 = 5$. The only combinations are $\{0,1,4,5\}$ and $\{0,2,3,5\}$. We can take $k=4$ to get $\{0,4,5,9\}$ and $k=2$ to get $\{0,4,6,10\}, both spanning sets.

Case 3: $d=7$, so $0=b_1 < b_2 < b_3 < b_4 = 5$. The only combination is $\{0,1,3,4\}$. We can take $k=5$ to get $\{0,4,5,9\}$, a spanning set.

And there is no way to have $d > 7$ and still fitting $b_2$ and $b_3$ without getting an arithmetic progression. So our proof is complete.

Product of all numbers

A prime $p$ is called "good" if there is an $m$ such that $p$ divides $m^2 - 2$. For example, 7 and 17 are good primes. Suppose $p$ is a good prime. Consider all numbers in the form of $a\sqrt{2} + b$ where $a,b \in \{0, \dots, p-1\}$ and $(a,b) \neq (0,0)$. The product of all those numbers can be written as $s\sqrt{2} + t$ where $s$ and $t$ are integers. Find their remainders modulo $p$. What if $p$ is not a good prime?

Solution

If $p$ is a good prime then $s \equiv t \equiv 0 \mod p$. If $p$ is not a good prime, then $s \equiv 0, t \equiv -1 \mod p$

Proof:

Define the "~" (equivalency) as follows: two numbers $s \sqrt{2} + t$ and $u\sqrt{2} + v$ are equivalent if $s \equiv u$ and $t \equiv v \mod p$. Then we readily see that the following properties are easy to show:

$X \sim X$ (reflexivity).

$X \sim Y \iff Y \sim X$ (symmetry).

If $X \sim Y$ and $Y \sim Z$ then $X \sim Z$ (transitivity).

Given these three then equivalency divides all numbers in te form $s\sqrt{2}+t$ into equivalency classes. Furthermore we see that:

If $X \sim Y$ and $W \sim Z$ then $X+W \sim Y+Z$ and $XW \sim YZ$. It follows directly from the properties of modular arithmetics.

Now, if $p$ is a good prime, then note that $(\sqrt{2} + m)(\sqrt{2} + p-m) = (2-m^2 + pm) + p\sqrt{2} \sim 0$. Thus the product of all numbers in question is equivalent to zero.

If $p$ is not a good prime, it's a lot more complicated.

First we claim the following: given $X \nsim 0$, then there exists $Y$ such that $XY \sim 1$. We call this $Y$ the inverse of $X$.

Proof: Given a number $a\sqrt{2} + b$, first we define the number $r$ as follows: $r \in (0, \dots, p-1)$ such that $r(b^2-2a^2) \equiv 1 \mod p$.

This $r$ must exist, because $b^2 - 2a^2 \not\equiv 0 \mod p$, for otherwise we have $(ba^{-1})^2 \equiv 2 \mod p$, a contradiction.

Then consider the number $ar\sqrt{2}-br$. The product of this and $a\sqrt{2} + b$ is: $$(ar\sqrt{2}-br)(a\sqrt{2} + b) = (2a^2-b^2)r \sim 1$$

Then we prove the second claim: If $XY \sim 0$ then either $X \sim 0$ or $Y \sim 0$.

Proof:

Let $X = a\sqrt{2}+b, Y = c\sqrt{2}+d$ then $XY = (bc+ad)\sqrt{2}+(2ac+bd) \sim 0$ so we have: $$bc+ad \equiv 0 \mod p$$ $$2ac+bd \equiv 0 \mod p$$ Assuming that both $X,Y$ are not equivalent to zero, either $c \neq 0$ or $d \neq 0$ because otherwise $Y \sim 0$. If $d \neq 0$ then: $$abc \equiv -a^2d \mod p$$ $$2abc \equiv - b^2d \mod p$$ $$\Rightarrow b^2d \equiv 2a^2d \mod p$$ $$(ba^{-1})^2 \equiv 2 \mod p$$ A contradiction. If $c \neq 0$ then we similarly show: $$abd \equiv - b^2c \mod p$$ $$abd \equiv -2a^2c \mod p$$ $$(ba^{-1})^2 \equiv 2 \mod p$$ Also a contradiction. (The assumption that $X \nsim 0$ is needed to take inverse of $a$)

So based on the second claim, we now assert that each number in the problem statement has an inverse, and the inverse is unique. For if $XY \sim XZ \sim 1$ then $X(Y-Z) \sim 0$ so $Y \sim Z$. This means that a lot of pairs among those numbers would multiply 1. The only numbers left are those who are inverses with itself. To find such numbers, note the following:

$$X^2 = 2ab\sqrt{2} + (2a^2+b^2) \sim 1$$ $$\iff ab \equiv 0, 2a^2+b^2 \equiv 1 \mod p$$ If $a \equiv 0$ then $b^2 \equiv 1$, so $p$ divides $b^2-1 = (b+1)(b-1)$. The only possible $b$ are $1, p-1$, and these numbers are in the problem statement.

If $b \equiv 0$ then $2a^2 \equiv 1$ which means $2 \equiv (a^{-1})^2 \mod p$, a contradiction.

So the only numbers who don't "cancel out" with other numbers are 1 and -1, and they're included in the final product. Thus the product of all those numbers is $\sim -1$.

Sequence modulo 2011

A sequence of integers $a_1, \dots, a_{2010}$ satisfy the following properties:

$a_1 - 1$ is divisible by 2011
$a_k a_{k-1} - k$ is divisible by 2011 for $k = 2, 3, \dots, 2010$

Show that $a_{2010} + 1$ is divisible by 2011.

Solution

First we prove the following assertion:

For $k$ odd, then $a_k.2.4.6. \cdots.(k-1) - 1.3.5.\cdots.k$ is divisible by 2011.
For $k$ even, then $a_k.1.3.5. \cdots .(k-1) - 2.4.6.\cdots.k$ is divisible by 2011.

Proof by induction. It can easily be seen for $k=1$ it's true. For $k=2$, we have $2011 | a_2a_1 - 2 = a_2(a_1-1) + (a_2-2)$ So $2011 | a_2 - 2$

Suppose it's true for $k-1$ odd, then for $k$ even:
$2011 | a_k a_{k-1} - k$ which means
$$2011 | 2.4.\cdots.(k-2) (a_k a_{k-1} - k) = 2.4.\cdots.(k-2)a_k a_{k-1} - 2.4.\cdots.(k-2).k$$
$$2011 | a_k (2.\cdots.(k-2)a_{k-1} - 1.3.\cdots.(k-1)) + (a_k.1.3.\cdots.(k-1) - 2.\cdots.(k-2).k)$$
And since 2011 divides the first term by induction hypothesis, then it also divides the second term, which completes the induction step. The proof for $k-1$ even and $k$ odd is similar.

Now, that means for $k=2010$, let $X = a_{2010}$ we have:
$$2011 | X.1.3. \cdots .2009 - 2.4.\cdots.2010$$

We now prove the following assertion for $i = 1,2,\dots,1005$
$$2011 | X.1.3. \cdots. (2011-2i) + (-1)^i (2i) (2i+2)\cdots.2010$$

For $i = 1$ it is true by definition. Now suppose it's true for $i-1$, then for $i > 1$:
$$2011 | X.1.3. \cdots. (2011 - 2i).(2011-2i+2) +(-1)^{i-1} (2i-2) (2i)\cdots.2010$$
if $2011 | a(2011+b) + c$ then $2011 | ab+c$
$$2011 | X.1.3. \cdots. (2011 - 2i).(-2i+2) + (-1)^{i-1} (2i-2) (2i)\cdots.2010$$
if $2011 | a$ then $2011 | -a$.
$$2011 | X.1.3. \cdots. (2011 - 2i).(2i-2) + (-1)^{i} (2i-2) (2i)\cdots.2010$$
$$2011 | 2(i-1)(X.1.3. \cdots. (2011 - 2i).+ (-1)^{i} (2i)\cdots.2010)$$
if $2011 | 2(i-1)a$ then $2011 | a$ since 2011 is prime and $i > 1$.

So the assertion holds for $i = 1,\dots, 1005$. Particularly, for $i = 1005$ we have:
$$2011 | X - 2010$$
$$2011 | X + 1 - 2011$$
$$2011 | X + 1$$

Number of solutions to modular equation

Prove that there are infinitely many prime numbers $p$ such that there are exactly $p^2$ integer triples $(x,y,z)$ such that $0 \leq x,y,z < p$ and $x^2+y^2 - 2010z^3$ is divisible by $p$.

Solution

We will first show that any prime that has form $p = 6k-1$ and does not divide 2010 satisfies the given condition.

First, let $p$ be such prime. Suppose there are $a,b \in \{ 1, \dots, p-1 \}$ such that $a^3 \equiv b^3 \mod p$.

Since $b^{p-1} \equiv 1 \mod p$ then let $c \equiv ab^{p-2} \mod p$, so that we have $bc \equiv a \mod p$.

Then $b^3c^3 \equiv a^3 \equiv b^3 \mod p$ which means $b^3c^3 \equiv b^3 \mod p$ which then means $c^3 \equiv 1 \mod p$ (since $(b,p) = 1$).

Because $c^{p-1} = c^{6k-2} \equiv 1 \mod p$ and $c^{6k-3} = c^{3(2k-1)} \equiv 1 \mod p$, then we must have $c \equiv 1 \mod p$, which means $a \equiv b \mod p$.

In summary, we've shown that $a^3 \equiv b^3 \Rightarrow a \equiv b \mod p$. This means that the set $\{1^3, \dots, p^3 \} \mod p$ is the same as set $\{ 1, \dots, p \} \mod p$. So given any arbitrary $x$ and $y$, we can find exactly one $z$ such that $z^3 \equiv 2010^{-1} (x^2+y^2) \mod p$. Since there are $p^2$ possible pairs for $(x,y)$, then there are also $p^2$ possible triples for $(x,y,z)$.

Now we show that there are infinitely many primes of the form $6k-1$ which would mean that there must be infinitely many primes of the form $6k-1$ that do not divide 2010.

Note that any prime above 3 must have form $6k+1$ or $6k-1$. If there are only finite number of primes of the form $6k-1$, say $p_1, p_2, \dots, p_n$, consider the number $N = 6p_1p_2 \dots p_n - 1$.
For each $i$, $p_i$ does not divide $N$ because otherwise $p_i$ would also have to divide 1, a contradiction. So $N$ is not divisible by any of the $p_i$s, which means all prime factors of $N$ must be of the form $6k+1$. That means, $N \equiv 1 \mod 6$, a contradiction.