Math Challenge - December 2021

[fresh_42]

Lemma: If $p$ is prime then the only solutions to $x^2 = 1 \text{ (mod p)}$ are $x = -1, 1$.

Proof: Suppose $x^2 = 1 \text{ (mod p)}$. Then $p \vert (x^2 - 1)$ i.e. $p \vert (x-1)(x+1)$. Since $p$ is prime, we have $p \vert x + 1$ or $p \vert x - 1$. This implies $x = -1 \text{ (mod p)}$ or $x = 1 \text{ (mod p)}$. And we see $1$ and $-1$ really are solutions to $x^2 = 1 \text{ (mod p)}$. []

Proof of problem in OP: $(\Longleftarrow):$ Suppose $p$ is prime. Then

$$(p-1)! = (p-1)(p-2)\cdot \dots \cdot 2 \cdot 1 = -1\cdot (p-2) \cdot \dots \cdot 2 \cdot 1 = -1 \cdot 1 = -1 \text{ (mod p)}$$

The second equality is from $p-1 = -1 \text{ (mod p)}$ and the third inequality is using the fact the only squares are $-1$ and $1$, therefore every other element is canceled by its inverse.

You should have mentioned the $p=2$ case, but ok, we can assume that $p$ is odd.

I don't get it. $(p-k)\cdot k=-k^2\in \{\pm 1\} \;(p)$. Why isn't it $-1$ an odd number of times?

$(\Longrightarrow):$ Suppose $(p-1)! = -1 \text{ (mod p)}$. We will show $p$ must be prime. Suppose $1 \le d < p$ is a divisor of $p$. Then $d$ is also a divisor of $(p-1)!$. We can rewrite the congruence as

$$-1 = (p-1)! + p\cdot k$$ for some integer $k$.

It would have been better to use $n$ instead of $p$. If you start with $p$ then your readers associate a prime whilst this is what you want to have at the end, and not at the beginning.

Since $d$ divides the right hand side, it must divide the left hand side. Hence, $d \vert -1$ which implies $d = 1$. This implies the only divisors of $p$ are $1$ and $p$. We may conclude $p$ is prime. []

Correct.

For the next part, we claim the first two primes that satisfy $(p-1)! = -1 \text{ (mod p^2)}$ are $p = 5$ and $p = 13$. Observe,

\begin{align}
(2-1)! &= 1 \text{ (mod 4)} \\
(3-1)! &= 2 \text{ (mod 9)} \\
(5-1)! &= 24 = -1 \text{ (mod 25)} \\
(7-1)! &= 720 = 34 \text{ (mod 49)} \\
(11-1)! &= 3628800 = 10 \text{ (mod 121)} \\
(13-1)! &= 479,001,600 = 168 = -1 \text{ (mod 169)}
\end{align}

Right. Up to now, there is only one more so-called Wilson prime known, namely $563.$ It is unknown whether there are more than that. If, then they are greater than $20,000,000,000,000.$ The conjecture is, that there are infinitely many Wilson primes.

 

[fishturtle1]

I don't get it. (p−k)⋅k=−k2∈{±1}(p). Why isn't it −1 an odd number of times?

Thank you for the feedback.

I think because the lemma shows for every element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$: either $g \in \lbrace \pm 1 \rbrace$ or $g \neq g^{-1}$. So $(p-1)! = -1\cdot1 \cdot1\cdot \dots \cdot 1 = -1$.

 

[fresh_42]

Thank you for the feedback.

I think because the lemma shows for every element $g \in (\mathbb{Z}/p\mathbb{Z})^\times$: either $g \in \lbrace \pm 1 \rbrace$ or $g \neq g^{-1}$. So $(p-1)! = -1\cdot1 \cdot1\cdot \dots \cdot 1 = -1$.

I still don't see it. It is a little bit more complicated, but you are on good track.

Hint: What can you say about $f(x):=x^p-x$?

 

[fishturtle1]

We can say $f(g) = g^p - g = g\cdot g^{p-1} - g = g \cdot 1 - g = 0$ for all $g \in (\mathbb{Z}/p\mathbb{Z})^\times$. So, $f(x)$ has $n$ distinct roots in $\mathbb{Z}/p\mathbb{Z}$. So we can factor it as $f(x) = (x-0)(x-1)(x-2)\cdot\dots\cdot (x-(p-1))$.Maybe something like this works: Consider the polynomial $f(x) = x^{p-1} - 1$. By Lagrange theorem, for every $g \in (\mathbb{Z}/p\mathbb{Z})^\times$, we have $f(g) = 0 \text{ (mod p)}$. So we can divide $f(x)$ by $x-k$ for $k = 1, 2, \dots, p-1$. So, we have $f(x) = \prod_{k=1}^{p-1} (x-k)$. Evaluating $f(x)$ at $x = 0$ gives

$$f(0) = 0^{p-1} - 1 = 0 -1 = -1 \text{ (mod p)}$$
and
$$f(0) = (-1)^{p-1} (p-1)! = (p-1)! \text{ (mod p)}$$

noting that $p-1$ is even. And so $(p-1)! = -1 \text{ (mod p)}$. []

 

[fresh_42]

We can say $f(g) = g^p - g = g\cdot g^{p-1} - g = g \cdot 1 - g = 0$ for all $g \in (\mathbb{Z}/p\mathbb{Z})^\times$. So, $f(x)$ has $n$ distinct roots in $\mathbb{Z}/p\mathbb{Z}$. So we can factor it as $f(x) = (x-0)(x-1)(x-2)\cdot\dots\cdot (x-(p-1))$.

Almost, i.e. modulo typos.

$x^p-x$ has at most $p$ roots in $\mathbb{Z}_p.$
$(x^p-x)'=px^{p-1}-1\equiv -1 \mod p$ so there are only simple roots.
(Otherwise, $f(x)$ and $f'(x)$ would have a common divisor!)
Since for all elements $x$ of the cyclic group $\mathbb{Z}^\times_p$ hold $x^{p-1}=1 \mod p$ there are at least $p$ roots of $(x-0)(x^{p-1}-1),$ namely the elements of $\mathbb{Z}_p$ if we consider $x^p-x\in \mathbb{Z}_p[x].$

Thus $f(x) = x^p-x=(x-1)(x-2)\cdot\dots\cdot (x-(p-1))(x-p).$

What now?

Edit: I didn't see your edit in time, sorry. I took
$$
\dfrac{f(x)}{x-p}(p)=(p-1)\cdots (p-(p-1))=(p-1)!=p^{p-1}-1=-1 \mod p
$$
but that is the same

 

[fishturtle1]

Thanks for your time and help!

 

[fishturtle1]

Definition: Let $G$ and $H$ be groups. We define $GH = \lbrace gh : g \in G, h \in H \rbrace$.

Definition: We say $G = N \ltimes H$ if $N$ is a normal subgroup of $G$, $H$ is a subgroup of $G$, $G = NH$ and $N \cap H = \lbrace 1 \rbrace$.

The group $\mathbb{Z}_2 \times \mathbb{Z}_3$ represents one group, up to isomorphism. However, we claim $\mathbb{Z}_{10}$ and $D_5$ are both the semi direct product $\mathbb{Z}_5 \ltimes \mathbb{Z}_2$, where $\mathbb{Z}_{10} \not\cong D_5$.

Proof: Consider the subgroups $N = \lbrace 0, 2, 4, 6, 8 \rbrace$ and $H = \lbrace 0, 5 \rbrace$ of $\mathbb{Z}_{10}$. Since $\mathbb{Z}_{10}$ is abelian, $N$ is normal in $\mathbb{Z}_{10}$. We see $N \cap H = \lbrace 0 \rbrace$. Lastly,

\begin{align}
2 + 5 &= 7 \\
4 + 5 &= 9 \\
6 + 5 &= 1 \\
8 + 5 &= 3 \\
\end{align}

Hence, $\mathbb{Z}_{10} = NH$. We can conclude $\mathbb{Z}_{10} = N \ltimes H$. (?)

Next, we write $D_5 = \langle a, b \vert a^5 = b^2 = 1, bab = a^{-1} \rangle$. Consider the subgroups $A = \lbrace 1, a, a^2, a^3, a^4 \rbrace$ and $B = \lbrace 1, b \rbrace$. Since $[D_5 : A] = 2$, we have $A$ is normal in $D_5$. We see $AB = \lbrace 1, a, a^2, a^3, a^4, b, ab, a^2b, a^3b, a^4b \rbrace = D_5$. We can conclude $D_5 = A \ltimes B$.

This shows groups that are not isomorphic can have the same semi direct product. []

Does it make sense to write $G = N \ltimes H$ since different groups can have the same semi direct product? Also, what is the difference between $\ltimes$ and $\rtimes$?

 

[fresh_42]

Definition: Let $G$ and $H$ be groups. We define $GH = \lbrace gh : g \in G, h \in H \rbrace$.

Definition: We say $G = N \ltimes H$ if $N$ is a normal subgroup of $G$, $H$ is a subgroup of $G$, $G = NH$ and $N \cap H = \lbrace 1 \rbrace$.

The other way around: $G=N \rtimes H= G$ if $N$ is the normal subgroup $N \trianglelefteq G$ and $H\leq G$ just a subgroup. A mnemonic is that you can cut $\rtimes$ in the middle and get $N \triangleleft G$ resp. $G \gt H$.

The group $\mathbb{Z}_2 \times \mathbb{Z}_3$ represents one group, up to isomorphism. However, we claim $\mathbb{Z}_{10}$ and $D_5$ are both the semi direct product $\mathbb{Z}_5 \ltimes \mathbb{Z}_2$, where $\mathbb{Z}_{10} \not\cong D_5$.

You recognized that I was asking about $\mathbb{Z}_2 \rtimes \mathbb{Z}_3$ and $\mathbb{Z}_2 \ltimes \mathbb{Z}_3$? What makes a product on the Cartesian set $\mathbb{Z}_2\times \mathbb{Z}_3$ direct, and what semidirect? What is the multiplication formula for semidirect products?

Proof: Consider the subgroups $N = \lbrace 0, 2, 4, 6, 8 \rbrace$ and $H = \lbrace 0, 5 \rbrace$ of $\mathbb{Z}_{10}$. Since $\mathbb{Z}_{10}$ is abelian, $N$ is normal in $\mathbb{Z}_{10}$. We see $N \cap H = \lbrace 0 \rbrace$. Lastly,

\begin{align}
2 + 5 &= 7 \\
4 + 5 &= 9 \\
6 + 5 &= 1 \\
8 + 5 &= 3 \\
\end{align}

Hence, $\mathbb{Z}_{10} = NH$. We can conclude $\mathbb{Z}_{10} = N \ltimes H$. (?)

All subgroups are normal in an abelian group, and all products are direct products.

Next, we write $D_5 = \langle a, b \vert a^5 = b^2 = 1, bab = a^{-1} \rangle$. Consider the subgroups $A = \lbrace 1, a, a^2, a^3, a^4 \rbrace$ and $B = \lbrace 1, b \rbrace$. Since $[D_5 : A] = 2$, we have $A$ is normal in $D_5$. We see $AB = \lbrace 1, a, a^2, a^3, a^4, b, ab, a^2b, a^3b, a^4b \rbrace = D_5$. We can conclude $D_5 = A \ltimes B$.

$D_5 = A \rtimes B$

This shows groups that are not isomorphic can have the same semi direct product. []

Does it make sense to write $G = N \ltimes H$ since different groups can have the same semi direct product? Also, what is the difference between $\ltimes$ and $\rtimes$?

see above

 

[fishturtle1]

You recognized that I was asking about Z2⋊Z3 and Z2⋉Z3? What makes a product on the Cartesian set Z2×Z3 direct, and what semidirect? What is the multiplication formula for semidirect products?

Definition: If $G$ and $H$ are groups, then the direct product $G \times H = \lbrace (g, h) : g \in G, h \in H \rbrace$ is the group whose operation is $(g_1, h_1) + (g_2, h_2) = (g_1 + g_2, h_1 + h_2)$.

If $G$ is a group and $N \rtimes H$ is a semi direct product of $G$, then $NH = \lbrace nh : n \in N, h \in H \rbrace = G$ where the multiplication $nh$ is in $G$.

Going back to the original problem: $\mathbb{Z}_6 = N \rtimes H = H \ltimes N$, where $N = \lbrace 0, 3 \rbrace \cong \mathbb{Z}_2$ and $H = \lbrace0, 2, 4 \rbrace \cong \mathbb{Z}_3$.

 

[fresh_42]

Definition: If $G$ and $H$ are groups, then the direct product $G \times H = \lbrace (g, h) : g \in G, h \in H \rbrace$ is the group whose operation is $(g_1, h_1) + (g_2, h_2) = (g_1 + g_2, h_1 + h_2)$.

If $G$ is a group and $N \rtimes H$ is a semi direct product of $G$, then $NH = \lbrace nh : n \in N, h \in H \rbrace = G$ where the multiplication $nh$ is in $G$.

This is not what makes a product semidirect.

... where the multiplication $nh$ is in $G$.

is wrong. $nh$ suggests the complex product $n\cdot h,$ but this is only true for direct products.
$$
(n_1h_1)\cdot (n_2h_2) =n_1h_1n_2h_2 \notin_\text{i.g.} NH
$$

Going back to the original problem: $\mathbb{Z}_6 = N \rtimes H = H \ltimes N$, where $N = \lbrace 0, 3 \rbrace \cong \mathbb{Z}_2$ and $H = \lbrace0, 2, 4 \rbrace \cong \mathbb{Z}_3$.

Again, $\mathbb{Z}_6$ is abelian and all products are direct since all subgroups are normal.

 

[fishturtle1]

I'm sorry i am confused. On the wikipedia page it says "As with direct products, there is a natural equivalence between inner and outer semi direct products, and both are commonly referred to simply as $\textit{semi direct products}$".

and

"Given any two groups $N$ and $H$ and a group homomorphism $\varphi : H \to Aut(N)$, we can construct a new group $N \rtimes_{\varphi} H$, called the outer semi direct product of $N$ and $H$ with respect to $\varphi$, defined as follows:

-The underlying set is the Cartesian product $N \times H$.
-The group operation is determined by the homomorphism $\varphi$: $(n_1, h_1) \cdot (n_2, h_2) = (n_1\varphi(h_1)(n_2), h_1h_2) = (n_1\varphi_{h_1}(n_2), h_1h_2)$ for all $n_1, n_2 \in N$ and $h_1, h_2 \in H$.

So is this what is meant by the operation on $N \rtimes H$? We have to specify $\varphi : H \to Aut(N)$ first?

 

[fresh_42]

I'm sorry i am confused. On the wikipedia page it says "As with direct products, there is a natural equivalence between inner and outer semi direct products, and both are commonly referred to simply as $\textit{semi direct products}$".

and

"Given any two groups $N$ and $H$ and a group homomorphism $\varphi : H \to Aut(N)$, we can construct a new group $N \rtimes_{\varphi} H$, called the outer semi direct product of $N$ and $H$ with respect to $\varphi$, defined as follows:

-The underlying set is the Cartesian product $N \times H$.
-The group operation is determined by the homomorphism $\varphi$: $(n_1, h_1) \cdot (n_2, h_2) = (n_1\varphi(h_1)(n_2), h_1h_2) = (n_1\varphi_{h_1}(n_2), h_1h_2)$ for all $n_1, n_2 \in N$ and $h_1, h_2 \in H$.

So is this what is meant by the operation on $N \rtimes H$? We have to specify $\varphi : H \to Aut(N)$ first?

Yes. That makes a big difference. The direct product belongs to $\varphi \equiv \operatorname{id}_N$ but any other homomorphism $\varphi$ defines another semidirect product. Now investigate the groups with six elements.

 

[fishturtle1]

We note that $Aut(\mathbb{Z}_3) = \lbrace id, \sigma \rbrace$ where $\sigma : 0 \mapsto 0, 1 \mapsto 2, 2 \mapsto 1$. This implies that the only homomorphisms from $\mathbb{Z}_2$ to $Aut(\mathbb{Z}_n)$ are $\tau$ and $\varphi$, where $\tau : 0, 1 \mapsto id$ and $\varphi: 0 \mapsto id, 1 \mapsto \sigma$.

We noted earlier that $\mathbb{Z}_3 \rtimes_\tau \mathbb{Z}_2 = \mathbb{Z}_3 \times \mathbb{Z}_2$. We will show $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2 \cong D_3$, by showing $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2$ is a group of $6$ elements and every element has order $\le 3$.

We note $(0,0)$ is the identity and has order $1$. Observe,

\begin{align}
(1,0)(1,0)(1,0) &= (1,0)(1 + \varphi(0)(1), 0+0) \\
&= (1,0)(1 + id(1), 0+0) \\
&= (1,0)(2,0) \\
&= (1 + \varphi(0)(2), 0+0) \\
&= (1 + id(2), 0+0) \\
&= (1+2, 0+0)\\
&= (0,0)
\end{align}

So, $(1,0)$ has order $3$. By a similar calculation, $(2,0)$ has order $3$. Next, we have

\begin{align}
(1,1)^2 &= (1 + \varphi(1)(1), 1+1) \\
&= (1 + \sigma(1), 1+1) \\
&= (1 + 2, 1+1) \\
&= (3, 2) \\
&= (0,0)
\end{align}

So, $(1,1)$ has order $2$. By a similar calculation, $(2,1)$ and $(0,1)$ have order $2$. We may conclude $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2 \cong D_3$. []

 

[fresh_42]

We note that $Aut(\mathbb{Z}_3) = \lbrace id, \sigma \rbrace$ where $\sigma : 0 \mapsto 0, 1 \mapsto 2, 2 \mapsto 1$. This implies that the only homomorphisms from $\mathbb{Z}_2$ to $Aut(\mathbb{Z}_n)$ are $\tau$ and $\varphi$, where $\tau : 0, 1 \mapsto id$ and $\varphi: 0 \mapsto id, 1 \mapsto \sigma$.

We noted earlier that $\mathbb{Z}_3 \rtimes_\tau \mathbb{Z}_2 = \mathbb{Z}_3 \times \mathbb{Z}_2$. We will show $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2 \cong D_3$, by showing $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2$ is a group of $6$ elements and every element has order $\le 3$.

We note $(0,0)$ is the identity and has order $1$. Observe,

\begin{align}
(1,0)(1,0)(1,0) &= (1,0)(1 + \varphi(0)(1), 0+0) \\
&= (1,0)(1 + id(1), 0+0) \\
&= (1,0)(2,0) \\
&= (1 + \varphi(0)(2), 0+0) \\
&= (1 + id(2), 0+0) \\
&= (1+2, 0+0)\\
&= (0,0)
\end{align}

So, $(1,0)$ has order $3$. By a similar calculation, $(2,0)$ has order $3$. Next, we have

\begin{align}
(1,1)^2 &= (1 + \varphi(1)(1), 1+1) \\
&= (1 + \sigma(1), 1+1) \\
&= (1 + 2, 1+1) \\
&= (3, 2) \\
&= (0,0)
\end{align}

So, $(1,1)$ has order $2$. By a similar calculation, $(2,1)$ and $(0,1)$ have order $2$. We may conclude $\mathbb{Z}_3 \rtimes_\varphi \mathbb{Z}_2 \cong D_3$. []

That's right, well done. It might have been a bit easier with $S_3\cong D_3$ and $A_3\cong \mathbb{Z}_3$ but this only shows that the same groups can be realized by different methods.

Do you have an opinion on $\mathbb{Z}_3\ltimes \mathbb{Z}_2$?

 

[fishturtle1]

We note that $Aut(\mathbb{Z}_2) = \lbrace id \rbrace$. So the only homomorphism $\varphi : \mathbb{Z}_3 \to Aut(\mathbb{Z}_2)$ is $\varphi: 0, 1, 2 \mapsto id$. And this implies $\mathbb{Z}_3 \ltimes_\varphi \mathbb{Z}_2 = \mathbb{Z}_3 \times \mathbb{Z}_2$. So in this case, we can write $\mathbb{Z}_3 \times \mathbb{Z}_2 = \mathbb{Z}_3 \ltimes \mathbb{Z}_2$?

 

[fresh_42]

We note that $Aut(\mathbb{Z}_2) = \lbrace id \rbrace$. So the only homomorphism $\varphi : \mathbb{Z}_3 \to Aut(\mathbb{Z}_2)$ is $\varphi: 0, 1, 2 \mapsto id$. And this implies $\mathbb{Z}_3 \ltimes_\varphi \mathbb{Z}_2 = \mathbb{Z}_3 \times \mathbb{Z}_2$. So in this case, we can write $\mathbb{Z}_3 \times \mathbb{Z}_2 = \mathbb{Z}_3 \ltimes \mathbb{Z}_2$?

Yes, or: $\mathbb{Z}_3 \ltimes_\varphi \mathbb{Z}_2$ makes no sense.

 

[QuantumSpace]

Why is 14 not solved yet?

Spoiler

I will assume $a \ne b$ and $X = \{a,b\}$.

All topologies on $X$ are
$$\{\emptyset, X \}, \{\emptyset, \{a\},X\}, \{\emptyset, \{b\}, X\}, \{\emptyset, \{a\}, \{b\}, X\}$$

Let us write $\tau_a = \{\emptyset, \{a\},X\}, \tau_b = \{\emptyset, \{b\},X\}$.

Then the map $f: (X, \tau_a)\to (X, \tau_b)$ given by $f(a) = b, f(b) = a$ is a homeomorphism since $f$ induces a bijection between the topologies (this is swiftly verified).

Thus $(X, \tau_a)\cong (X,\tau_b)$. There are no other homeomorphisms, because the topologies do not have the same cardinality.

Finally, for the indiscrete topology $\{\emptyset, X\}$ every net converges. Indeed, if $\{x_i\}_{i \in I}\subseteq \{a,b\}$ is a net, it converges to both $a$ and $b$. This is immediate from the definition of convergence:
$$x_i \to x \iff \forall V \in \mathcal{V}(x): \exists i_0 \in I: \forall i \ge i_0: x_i \in V$$
In this case, $\mathcal{V}(x) = \{X\}$ and this satisfied trivially. More generally, we can take any set $X$ with the indiscrete topology and all nets in it will converge to every point in the entire space.

 

[julian]

19.

Spoiler

We have

\begin{align*}
P_n = 2 P_{n-1} + P_{n-2} , \qquad \text{for } \; n \geq 2,
\end{align*}

with $P_0 = 0$ and $P_1 = 1$ ($P_2 = 2$).

Write

\begin{align*}
F(x) = \sum_{k=0}^\infty P_k x^k
\end{align*}

Then

\begin{align*}
F(x) & = P_1 x + P_2 x^2 + P_3 x^3 + P_4 x^4 + \cdots
\nonumber \\
2x F(x) & = \qquad \; 2 P_1 x^2 + 2P_2 x^3 + 2P_3 x^4 + \cdots
\nonumber \\
x^2 F(x) & = \qquad \qquad \qquad \quad P_1 x^3 + P_2 x^4 + \cdots
\end{align*}

and so

\begin{align*}
F(x) = 2 x F(x) + x^2 F(x) + x
\end{align*}

then

\begin{align*}
F(x) & = \frac{x}{1 - 2x - x^2}
\nonumber \\\
& = \frac{1}{2 \sqrt{2}} \left( \frac{1}{1- (1 + \sqrt{2}) x} - \frac{1}{1- (1 - \sqrt{2}) x} \right)
\nonumber \\\
& = \frac{1}{2 \sqrt{2}} \sum_{n=0}^\infty \left( (1 + \sqrt{2})^n - (1 - \sqrt{2})^n \right) x^n
\end{align*}

We read off that

\begin{align*}
P_n = \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2 \sqrt{2}}
\end{align*}

 

[fresh_42]

19.

Spoiler

We have

\begin{align*}
P_n = 2 P_{n-1} + P_{n-2} , \qquad \text{for } \; n \geq 2,
\end{align*}

with $P_0 = 0$ and $P_1 = 1$ ($P_2 = 2$).

Write

\begin{align*}
F(x) = \sum_{k=0}^\infty P_k x^k
\end{align*}

Then

\begin{align*}
F(x) & = P_1 x + P_2 x^2 + P_3 x^3 + P_4 x^4 + \cdots
\nonumber \\
2x F(x) & = \qquad \; 2 P_1 x^2 + 2P_2 x^3 + 2P_3 x^4 + \cdots
\nonumber \\
x^2 F(x) & = \qquad \qquad \qquad \quad P_1 x^3 + P_2 x^4 + \cdots
\end{align*}

and so

\begin{align*}
F(x) = 2 x F(x) + x^2 F(x) + x
\end{align*}

then

\begin{align*}
F(x) & = \frac{x}{1 - 2x - x^2}
\nonumber \\\
& = \frac{1}{2 \sqrt{2}} \left( \frac{1}{1- (1 + \sqrt{2}) x} - \frac{1}{1- (1 - \sqrt{2}) x} \right)
\nonumber \\\
& = \frac{1}{2 \sqrt{2}} \sum_{n=0}^\infty \left( (1 + \sqrt{2})^n - (1 - \sqrt{2})^n \right) x^n
\end{align*}

We read off that

\begin{align*}
P_n = \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2 \sqrt{2}}
\end{align*}

It is important to notice that your proof takes place entirely in $\mathbb{R}[[x]],$ i.e. the ring of formal real power series. $F(x)$ is neither a power series of a function nor is it a quotient of polynomials. All there is, is $\mathbb{R}^{\mathbb{N}_0}.$

One can give a purely analytical proof by the observation
\begin{align*}
L:=\lim_{n \to \infty}\dfrac{P_{n}}{P_{n-1}}&=2+\lim_{n \to \infty}\dfrac{P_{n-2}}{P_{n-1}}=2+\dfrac{1}{L}
\end{align*}
In that case, we find the same result but would have to show that $L$ actually exists.

$(P_n)_{n\in \mathbb{N}_0}$ is called Pell sequence and $L=1+\sqrt{2}$ the silver ratio.

 

[Fred Wright]

Problem #2
We have,
$$
I=\int_0^1 \frac{\log(x)}{z_1-x}dx- \int_0^1 \frac{\log(x)}{z_0+x}dx=I_1 - I_2
$$
where
$$
z_0=|a|e^{i\phi}
$$
$$
|a| < 1
$$
$$
z_1=z_0+1
$$
Integrate $I_1$ by parts:
$$
I_1=\int_0^{1}\frac{\log{x}}{z_1-x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \int_0^{1}\frac{\log(z_1- x)}{x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \int_0^{1}\frac{\log(z_1)}{x}dx + \int_0^{1}\frac{\log(1-\frac{x}{z_1})}{x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \log(z_1)\log(x)|_0^1 +\int_0^{\frac{1}{z_1}}\frac{\log(1-u)}{u}du
$$
where we have made the substitution $u=\frac{x}{z_1}$ in the last term. The surface terms cancel and we are left with
$$
I_1=-Li_2(\frac{1}{z_1})
$$
where $Li_2(z)$ is the dilogarithm. In like manner we integrate $I_2$ by parts; again the surface terms cancel and making the substitution $u=-\frac{x}{z_0}$ we are left with,
$$
I_2=-Li_2(-\frac{1}{z_0})
$$
and thus,
$$
I=-(Li_2(\frac{1}{z_1})+Li_2(-\frac{1}{z_0}))=-(Li_2(\frac{1}{z_1})+Li_2(\frac{1}{1-z_1}))
$$
We make use of the property of the dilogarithm function:
$$
Li_2(1-z) + Li_2(1-\frac{1}{z})=-\frac{1}{2}\log^2(z)
$$
by making the change of variables,
$$
\frac{1}{z_1}=1-z
$$
$$
\frac{1}{1-z_1}=1-\frac{1}{z}
$$
to find,
$$
z=\frac{z_0}{z_1}
$$
$$
z=(\frac{|a|}{\sqrt{|a|^2+2\cos(\phi)|a|+1}})e^{i\tan^{-1}(\frac{|a|\sin(\phi)}{|a|+\cos(\phi)})}
$$
Thus,
$$
I=\frac{1}{2}(\log{(\frac{|a|}{\sqrt{|a|^2+2\cos(\phi)|a|+1}})}+i\tan^{-1}(\frac{|a|\sin(\phi)}{|a|+\cos(\phi)}))^2
$$

 

[fresh_42]

Problem #2
We have,
$$
I=\int_0^1 \frac{\log(x)}{z_1-x}dx- \int_0^1 \frac{\log(x)}{z_0+x}dx=I_1 - I_2
$$
where
$$
z_0=|a|e^{i\phi}
$$
$$
|a| < 1
$$
$$
z_1=z_0+1
$$
Integrate $I_1$ by parts:
$$
I_1=\int_0^{1}\frac{\log{x}}{z_1-x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \int_0^{1}\frac{\log(z_1- x)}{x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \int_0^{1}\frac{\log(z_1)}{x}dx + \int_0^{1}\frac{\log(1-\frac{x}{z_1})}{x}dx
$$
$$
=-\log(z_1 -x)\log(x)|_0^1 + \log(z_1)\log(x)|_0^1 +\int_0^{\frac{1}{z_1}}\frac{\log(1-u)}{u}du
$$
where we have made the substitution $u=\frac{x}{z_1}$ in the last term. The surface terms cancel and we are left with
$$
I_1=-Li_2(\frac{1}{z_1})
$$
where $Li_2(z)$ is the dilogarithm. In like manner we integrate $I_2$ by parts; again the surface terms cancel and making the substitution $u=-\frac{x}{z_0}$ we are left with,
$$
I_2=-Li_2(-\frac{1}{z_0})
$$
and thus,
$$
I=-(Li_2(\frac{1}{z_1})+Li_2(-\frac{1}{z_0}))=-(Li_2(\frac{1}{z_1})+Li_2(\frac{1}{1-z_1}))
$$
We make use of the property of the dilogarithm function:
$$
Li_2(1-z) + Li_2(1-\frac{1}{z})=-\frac{1}{2}\log^2(z)
$$
by making the change of variables,
$$
\frac{1}{z_1}=1-z
$$
$$
\frac{1}{1-z_1}=1-\frac{1}{z}
$$
to find,
$$
z=\frac{z_0}{z_1}
$$
$$
z=(\frac{|a|}{\sqrt{|a|^2+2\cos(\phi)|a|+1}})e^{i\tan^{-1}(\frac{|a|\sin(\phi)}{|a|+\cos(\phi)})}
$$
Thus,
$$
I=\frac{1}{2}(\log{(\frac{|a|}{\sqrt{|a|^2+2\cos(\phi)|a|+1}})}+i\tan^{-1}(\frac{|a|\sin(\phi)}{|a|+\cos(\phi)}))^2
$$

Oh dear, what have you done? The first term is at least near the solution (more or less), however, I do not have the second. But that might be due to the fact that my solution doesn't split into real and imaginary parts. It doesn't use complex numbers at all. The solution isn't that complicated, only a square, a logarithm, and a root.

The integral itself is rather tricky and uses two auxiliary functions which are not evident and a double integration.

 

[Fred Wright]

Oh dear, what have you done? The first term is at least near the solution (more or less), however, I do not have the second. But that might be due to the fact that my solution doesn't split into real and imaginary parts. It doesn't use complex numbers at all. The solution isn't that complicated, only a square, a logarithm, and a root.

The integral itself is rather tricky and uses two auxiliary functions which are not evident and a double integration.

From the problem statement, I assumed that $a$ is complex.

 

[fresh_42]

From the problem statement, I assumed that $a$ is complex.

Yes, but it remains $a$ throughout the calculation.

Hint: Consider
$$
F(x):=\dfrac{x\log x}{a+x}-\log(a+x)
$$
and
$$
G(x):=\dfrac{x\log x}{a+1-x}+\log(a+1-x)
$$

 

[TeethWhitener]

First of all, I wanted to thank @fresh_42 for all the work he’s done on these challenges. I don’t understand very many of them, but it’s a lot of fun and very educational to watch people tackle these problems (and occasionally to tackle a few myself).

Alright, since there are a few easy ones left, I’ll take a crack at those.

Spoiler: Problem 6

As far as I can tell, there are two approaches: 1) simply count the number of primes less than $7808000000$, or 2) we’re supposed to use the prime number theorem:
$$\pi(n)\approx\frac{n}{\ln{n}}$$
I’m guessing since the population is approximate, the answer should be approximate too. I’m also on winter break right now and am too lazy to write a script that actually counts the primes, so I’m going with method 2, which gives
$$\pi(7808000000)\approx342780659$$
We can get a little more accurate using the logarithmic integral:
$$Li(7808000000)=\int_{2}^{7808000000}{\frac{1}{\ln{x}}\text{dx}}=359364372$$

Spoiler: Problem 11

This is a straightforward calculation of the matrix exponential:
$$\exp{(H)}=\sum_{k=0}^{\infty}{\frac{1}{k!}H^k}$$
Mercifully, this series terminates rather quickly:
$$H^0=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$$
$$H^1=
\begin{pmatrix}
0 & x_1 & x_3 \\
0 & 0 & x_2 \\
0 & 0 & 0
\end{pmatrix}$$
$$H^2=
\begin{pmatrix}
0 & 0 & x_1x_2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
All matrices $H^3$ and higher are the zero matrix. Add together the terms that survive and we have:
$$\exp{(H)}=
\begin{pmatrix}
1 & x_1 & x_3+\frac{x_1x_2}{2} \\
0 & 1 & x_2 \\
0 & 0 & 1
\end{pmatrix}$$

 

[fresh_42]

First of all, I wanted to thank @fresh_42 for all the work he’s done on these challenges. I don’t understand very many of them, but it’s a lot of fun and very educational to watch people tackle these problems (and occasionally to tackle a few myself).

Alright, since there are a few easy ones left, I’ll take a crack at those.

Spoiler: Problem 6

As far as I can tell, there are two approaches: 1) simply count the number of primes less than $7808000000$, or 2) we’re supposed to use the prime number theorem:
$$\pi(n)\approx\frac{n}{\ln{n}}$$
I’m guessing since the population is approximate, the answer should be approximate too. I’m also on winter break right now and am too lazy to write a script that actually counts the primes, so I’m going with method 2, which gives
$$\pi(7808000000)\approx342780659$$
We can get a little more accurate using the logarithmic integral:
$$Li(7808000000)=\int_{2}^{7808000000}{\frac{1}{\ln{x}}\text{dx}}=359364372$$

Spoiler: Problem 11

This is a straightforward calculation of the matrix exponential:
$$\exp{(H)}=\sum_{k=0}^{\infty}{\frac{1}{k!}H^k}$$
Mercifully, this series terminates rather quickly:
$$H^0=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$$
$$H^1=
\begin{pmatrix}
0 & x_1 & x_3 \\
0 & 0 & x_2 \\
0 & 0 & 0
\end{pmatrix}$$
$$H^2=
\begin{pmatrix}
0 & 0 & x_1x_2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
All matrices $H^3$ and higher are the zero matrix. Add together the terms that survive and we have:
$$\exp{(H)}=
\begin{pmatrix}
1 & x_1 & x_3+\frac{x_1x_2}{2} \\
0 & 1 & x_2 \\
0 & 0 & 1
\end{pmatrix}$$

Yes. If we enumerate all humans on Earth, then a little more than there are US citizens get prime numbers.
$$
\exp{(H)}=
\begin{pmatrix}
1 & x_1 & x_3+\frac{x_1x_2}{2} \\
0 & 1 & x_2 \\
0 & 0 & 1
\end{pmatrix} \in \begin{bmatrix}
1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1
\end{bmatrix}$$
is the Heisenberg group.

The application that led Hermann Weyl to an explicit realization of the Heisenberg group was the question of why the Schrödinger picture and Heisenberg picture are physically equivalent.

 

[fishturtle1]

Spoiler: Problem 17

First we will show $n$ contains no squares. Suppose $p \vert n$. Then $p \vert \left(\frac{n}{p }- 1\right)$. Then there exists integers $k, l$ such that $n = pk$ and $\frac{n}{p} - 1 = pl$. Multiplying the second equation by $p$ on both sides gives $n - p = p^2l$. Rearranging gives $n = p(pl + 1)$. We see that $p$ does not divide $pl + 1$. We can conclude $p \vert n$, but $p^e$ does not divide $n$ for $e \ge 2$. This shows that $n$ is square free.

Next, assume by contradiction $n = pq$ for distinct primes $p$ and $q$. Since $p \vert n$, we have $p \vert \left(\frac{n}{p}- 1\right)$. But this implies $p \vert (q - 1)$. By similar reasoning, $q \vert (p-1)$. So, $$p \le q - 1 \le q \le p -1$$ which is a contradiction. We may conclude $n$ cannot be a semi prime.

 

[fresh_42]

Spoiler: Problem 17

First we will show $n$ contains no squares. Suppose $p \vert n$. Then $p \vert \left(\frac{n}{p }- 1\right)$. Then there exists integers $k, l$ such that $n = pk$ and $\frac{n}{p} - 1 = pl$. Multiplying the second equation by $p$ on both sides gives $n - p = p^2l$. Rearranging gives $n = p(pl + 1)$. We see that $p$ does not divide $pl + 1$. We can conclude $p \vert n$, but $p^e$ does not divide $n$ for $e \ge 2$. This shows that $n$ is square free.

Next, assume by contradiction $n = pq$ for distinct primes $p$ and $q$. Since $p \vert n$, we have $p \vert \left(\frac{n}{p}- 1\right)$. But this implies $p \vert (q - 1)$. By similar reasoning, $q \vert (p-1)$. So, $$p \le q - 1 \le q \le p -1$$ which is a contradiction. We may conclude $n$ cannot be a semi prime.

Correct.

Numbers with those properties are called Giuga numbers. Giuga conjectured $1950$ that a natural number is prime, if and only if
$$
\sum_{k=1}^{n-1}k^{n-1}\equiv -1 \mod n.
$$
The equation follows from Fermat's little theorem if $n$ is prime:
$$
k^{p-1}\equiv 1\mod p \Longrightarrow \sum_{k=1}^{n-1}k^{n-1}\equiv (p-1)\cdot 1=p-1 \equiv -1 \mod p
$$
It is not clear whether the other direction holds, i.e. whether there are composite numbers with this property. It is only known that such a number has at least $10,000$ decimal digits. A Carmichael number $n$ is a composite, square-free number with the additional property
$$
a^{n-1}\equiv 1 \mod n \text{ for all coprime } a \; , \;(a,n)=1.
$$
Korselt has shown $1899$ that a number $n$ is a Carmichael number, if it is not prime, square-free, and for all its prime divisors $p\,|\,n$ holds $(p-1)\,|\,(n-1).$ This result can be tightened to
$$
p\,|\,n\Longrightarrow (p-1)\,|\,\left(\dfrac{n}{p}-1\right)
$$
because $n-1=(n/p)-1+(p-1)(n/p),$ i.e. $n-1\equiv (n/p)-1\mod (p-1).$ This shows, that Carmichael numbers and Giuga numbers are closely related. Giuga's conjecture is indeed equivalent to:

No natural number is simultaneously Giuga and Carmichael number.

Another interesting theorem is, that $n$ is a Giuga number, if and only if it is a composite, square-free number and
$$
\sum_{p|n}\dfrac{1}{p} -\prod_{p|n}\dfrac{1}{p} \in \mathbb{N}.
$$
This term equals $1$ for all known Giuga numbers:
\begin{align*}
&30, 858, 1722, 66198, 2214408306, 24423128562, 432749205173838, \\&14737133470010574, 550843391309130318,\\& 244197000982499715087866346,\\& 554079914617070801288578559178,\\& 1910667181420507984555759916338506
\end{align*}

 

[PeroK]

Here's an exact answer to problem 6.

Spoiler: Problem 6

I wrote a Python script to count the prime numbers up to 10 billion. For a population of $7,808,000,000$, the exact number of primes less than this is:$$359,357,713$$

Taking the current population to be about $7,916,000,000$, then the exact number of primes less than this is: $$364,098,533$$ That said, there is a website that gives you the precise value of $\pi(n)$ for large value of $n$:

https://www.dcode.fr/prime-number-pi-count

 

[julian]

Quick go at 27.

Spoiler

As $f(x)$ is a continuous function on $[a,b]$ we have that it is bounded and attains those bounds:

\begin{align*}
m \leq f(x) \leq M
\end{align*}

for $x \in [a,b]$, where $m$ is the lower bound of $f(x)$ and $M$ is the upper bound of $f(x)$. So that

\begin{align*}
m \int_a^b g (x) \leq \int_a^b f(x) g (x) dx \leq M \int_a^b g (x)
\end{align*}

where we have used that $g(x) \geq 0$ and that $\int_a^b g(x)$ is finite (as $g(x)$ is integrable). Thus there is a number, say $N$, such that $m \leq N \leq M$ and

\begin{align*}
\int_a^b f(x) g (x) dx = N \int_a^b g (x) .
\end{align*}

As $f(x)$ is continuous on $[a,b]$ the function takes every value between $f(a)$ and $f(b)$, thus there exists $m \leq f(\xi) \leq M$ ($f(\xi)=N$) such that

\begin{align*}
\int_a^b f(x) g (x) dx = f (\xi) \int_a^b g(x) dx
\end{align*}

 

[TeethWhitener]

Spoiler: Problem 23

Edit: ignore what I wrote. I didn’t consider that the number could be prime all by itself. The first two conditions ($2\mod3$ and $3\mod4$) tell us that the solution will be an integer of the form $12n-1$. The first and third ($2\mod5$) conditions tell us that it ends in a 7. The smallest number that satisfies this is 47.

Pretty sure a little piece of code could do this much quicker, but here’s my calculator-free solution:
The smallest prime factor will be at least 7. Looking at the modular behavior of multiples of 7, we find that a family of solutions consisting of 7 multiplied by a number satisfying the following expressions simultaneously:
$$3n+2$$
$$4n+1$$
$$5n+1$$
The smallest number of this form is 41 (this is easy to find by noting that $4n+1$ and $5n+1$ together imply $20n+1$), which gives a solution of $7\times41=287$.

To check whether this is the smallest integer satisfying the equations, we note that $287<\sqrt{17}$, so the smallest prime factor of the solution will either be 7, 11, or 13. The requirements for the multipliers of 11 are:
$$3n+1$$
$$4n+1$$
$$5n+2$$
The smallest number satisfying these expressions is 37 (again, easy to find noting that $3n+1$ and $4n+1$ together imply $12n+1$), which gives the solution $11\times37=407>287$.
The requirements for multipliers of 13 are:
$$3n+2$$
$$4n+3$$
$$5n+4$$
The smallest number satisfying these expressions is 59 (the conditions are equivalent to $3n-1$, $4n-1$, and $5n-1$ together, or $60n-1$), giving $13\times59=767>287$, so our answer is in fact 287.

 

[berkeman]

I'm posting the next two challenges for @fresh_42 since he is away from the Internet for the next couple of days:

28. Consider the circle segment above $A=(-1,0)$ and $B=(1,0)$ of
$$
x^2+\left(y+\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{4}{3}\,.
$$
The point $P:=\left(\dfrac{1}{\sqrt{3}},1-\dfrac{1}{\sqrt{3}}\right)$ lies on this segment. Calculate the height $h$ of the circle segment, and $|AP|+|PB|.$

29. Let $\varphi :V\longrightarrow V$ a linear mapping. Prove
$$
\operatorname{ker} (\varphi) \cap \operatorname{im}(\varphi) =\{0\} \Longleftrightarrow \operatorname{ker}(\varphi \circ \varphi )=\operatorname{ker}(\varphi )
$$

 

[ergospherical]

For 29; I can’t write latex at the moment so I’ll use f instead of phi.

In forward direction:
If v is in ker(f.f) then f(fv) = 0, therefore fv is in both ker(f) and im(f), thus fv = 0 and v is in ker(f). Similarly, if v is in ker(f) then fv = 0, therefore f(fv) = f(0) = 0 since 0 is in ker(f), and therefore v is also in ker(f.f). Hence ker(f.f) = ker(f).

In the opposite direction:
if ker(f.f) = ker(f), then fv = 0 implies that f(fv) = f(0) = 0, and therefore 0 is in both ker(f) and im(f). Consider some non-zero element q = fw of im(f), assuming one exists at all. Then fq = f(fw). But if fq = f(fw) = 0 were true, then w would be an element of ker(f.f) and therefore also of ker(f), which would imply that q = fw = 0. Therefore, 0 is the only element common to ker(f) and im(f).

 

[PeroK]

For 29; I can’t write latex at the moment so I’ll use f instead of phi.

In forward direction:
If v is in ker(f.f) then f(fv) = 0, therefore fv is in both ker(f) and im(f), thus fv = 0 and v is in ker(f). Similarly, if v is in ker(f) then fv = 0, therefore f(fv) = f(0) = 0 since 0 is in ker(f), and therefore v is also in ker(f.f). Hence ker(f.f) = ker(f).

In the opposite direction:
if ker(f.f) = ker(f), then fv = 0 implies that f(fv) = f(0) = 0, and therefore 0 is in both ker(f) and im(f). Consider some non-zero element q = fw of im(f), assuming one exists at all. Then fq = f(fw). But if fq = f(fw) = 0 were true, then w would be an element of ker(f.f) and therefore also of ker(f), which would imply that q = fw = 0. Therefore, 0 is the only element common to ker(f) and im(f).

I think that hangs together. Alternatively:

First note that, as $\phi(v) = 0 \ \Rightarrow \phi(\phi(v)) = 0$, we always have:
$$\ker(\phi) \subseteq \ker(\phi \circ \phi) \ \ (1)$$
Suppose $\ker(\phi) \cap im(\phi) = \{0\}$.

Let $v \in \ker(\phi \circ \phi)$ and $w = \phi(v)$. Note that $\phi(w) = 0$.

We have $w \in im(\phi)$ and $w \in ker(\phi)$, hence $w \in \ker(\phi) \cap im(\phi)$.

$\therefore w = 0$ and $v \in ker(\phi)$

$\therefore \ker(\phi \circ \phi)\subseteq \ker(\phi)$, and using equation (1) we have $\ker(\phi \circ \phi) = \ker(\phi)$.

Suppose $\ker(\phi \circ \phi) = \ker(\phi)$.

Let $v \in \ker(\phi) \cap im(\phi)$. Then $\phi(v) = 0$ and $v = \phi(u)$, some $u \in V$.

$\therefore \phi(\phi(u)) = 0, \ \text{hence} \ u \in ker(\phi \circ \phi)$

$\therefore u \in ker(\phi), \ \text{hence} \ v = \phi(u) = 0$

$\therefore \ker(\phi) \cap im(\phi) = \{0\}$

 

[fishturtle1]

Spoiler: Problem 22

We consider the first map first. Suppose $f(a) = f(b)$. Then $\lbrace a \rbrace =\lbrace b \rbrace$. This implies $a = b$. So, $f$ is injective. If $\vert X \vert = 1$, then $f$ is surjective. Suppose $\vert X \vert \ge 2$. Then we can write $X = \lbrace a, b, \dots \rbrace$. Then $\lbrace a, b \rbrace \in \mathcal{P}(X)$ and $f(x) \neq \lbrace a, b \rbrace$ because $\vert f(x) \vert = 1$ for all $x \in X$. So, $f$ is surjective iff $\vert X \vert = 1$. Lastly, we see $f^{-1}(\emptyset) = \lbrace x \in X : f(x) = \emptyset \rbrace = \emptyset$ because $\vert f(x) \vert = 1$ for all $x \in X$.

Next, we consider the second map. Let $X =\lbrace a ,\dots \rbrace$ and define $A = \lbrace a \rbrace$ and $B = \emptyset$. Then $$g((A,B)) = A \cup B = \lbrace a \rbrace \cup \emptyset = \lbrace a \rbrace = \emptyset \cup \lbrace a \rbrace = B \cup A = g((B,A))$$

and $(A,B) \neq (B,A)$. This shows $g$ is not injective. Let $C \in \mathcal{P}(X)$. Then $g((C,\emptyset)) = C \cup \emptyset = C$. This shows $g$ is surjective. Lastly, observe for arbitrary $A, B \in \mathcal{P}(X)$, if $A \cup B = \emptyset$, then $A \subseteq \emptyset$ and $B \subseteq \emptyset$. This implies $A = \emptyset$ and $B = \emptyset$. So, $g^{-1}(\emptyset) = \lbrace (\emptyset, \emptyset) \rbrace$. []

 

[mathwonk]

for #9, I believe that a loop is null homologous iff every holomorphic one - form integrates to zero over it. does that do it? (just throwing out an approach for complex analysts, since a topologist presumably knows that homology is a continuous invariant; or does 0 - homologue have some other meaning?)