Challenge Math Challenge - April 2021

[Office_Shredder]

For number 10

Spoiler

short proof:
The hessian matrix is symmetric with real eigenvalues. There's a theorem that if all its eigenvalues are negative then you have a local maximum, if they are all positive then it's a local minimum, and if there's at least one of each sign then it's neither a minimum or a maximum.

Since the dimension of the space is even, the determinant is the product of an even number of real numbers, and since that gives us a negative result there must be at least one positive and one negative eigenvalue, and hence $a$ is neither a maximum or a minimum

Here's a full proof with details, using Taylor polynomials in multiple dimensions:we write down the second degree Taylor polynomial with its error term. $(Hf)(a)$ is the $2n\times 2n$ matrix of second degree partial derivatives at $a$, and $(Df)(a)$ is a $1\times 2n$ row vector of the first order partial derivatives. $x$ and $a$ are column vectors, and everything below is regular matrix multiplication. Then a generally true thing for twice continuously differentiable functions

$$f(x)= f(a)+\left((Df)(a)\right)(x-a) + (x-a)^t\left((Hf)(a)\right)(x-a)+ (x-a)^t h(x)(x-a)$$

Where $h(x)$ is a matrix with entries $h_{ij}(x)$ that are continuous and $\lim_{x\to a}h_{ij}(x)=0$.

For the specific situation in #10, $(Df)(a)=0$, and the determinant of $(Hf)(a)$ is negative. Since $(Hf)(a)$ is a symmetric matrix, its eigenvalues are all real numbers. Furthermore since it's symmetric it is diagonalizable with the diagonal entries being the eigenvalues. There are an even number of them, so for the product of them to be a negative number there must be both a positive and a negative eigenvalue.

Suppose $v$ is a unit eigenvector with eigenvalue $\lambda$ and consider $f(a+\epsilon v)$ for $\epsilon >0$. Let's also write $H=(Hf)(a)$. Then we get
$$f(a+\epsilon v) = f(a) + \epsilon^2 v^t H v + \epsilon^2 v^t h(a+\epsilon v) v = f(a) + \epsilon^2 \left(\lambda+v^t h(a+\epsilon v) v\right)$$.

We will now show that if $\epsilon$ is small enough, $|v^t h(a+\epsilon v) v| < |\lambda/2|$. In general for a matrix $M$ whose largest entry is $\kappa$, and for any unit vector $v$,

$$|v^t M v| = |\sum_{i,j} M_{ij} v_i v_j | \leq \sum_{ij} |M_{ij} v_i v_j|$$

We use the fact that $|M_{ij}|\leq \kappa$ and $|v_k|\leq 1|$ for all $k$ since $v$ is a unit vector (obviously this is a crude bound, I think you can probably do $\sqrt{n}$ better but we don't care) to get (remember $M$ is $2n\times 2n$)
$$|v^t M v| \leq \sum_{i,j} \kappa = 4n^2 \kappa$$.

We know that the entries of the error term $h(a+\epsilon v)$ go to 0 as $\epsilon$ goes to zero, so if we pick $\epsilon$ small enough that all its entries are smaller in magnitude than $\frac{|\lambda|}{8n^2}$, and then $|v^t h(a+\epsilon v) v| \leq \lambda/2$. Let's let $g(\epsilon)= v^t h(a+\epsilon v) v$, and restrict ourselves to $\epsilon$ small enough that $|g(\epsilon)| \leq \lambda/2$.So we have $f(a+\epsilon v)= f(a)+\epsilon^2(\lambda +g(\epsilon))$. If we pick $v$ to be an eigenvector with positive eigenvalue, then $f(a+\epsilon v) \geq f(a) + \epsilon^2 \lambda/2 > f(a)$. If we pick $v$ to be an eigenvector with a negative eigenvalue , which I'll write as $-\lambda$ for $\lambda > 0$, then $f(a+\epsilon v) \leq f(a) - \epsilon^2 \lambda/2 < f(a)$. So $a$ is neither a local maximum or a local minimum.

 

[fresh_42]

For number 10

Spoiler

short proof:
The hessian matrix is symmetric with real eigenvalues. There's a theorem that if all its eigenvalues are negative then you have a local maximum, if they are all positive then it's a local minimum, and if there's at least one of each sign then it's neither a minimum or a maximum.

Since the dimension of the space is even, the determinant is the product of an even number of real numbers, and since that gives us a negative result there must be at least one positive and one negative eigenvalue, and hence $a$ is neither a maximum or a minimum

It's also short if we use the integral remainder of the Taylor seriess:
$$
f(\vec{a}+\vec{h})= f(\vec{a})+\int_0^1(1-t)\langle \vec{h},Hf(\vec{a}+t\vec{h})\vec{h} \rangle\,dt
$$

 

[Office_Shredder]

It's also short if we use the integral remainder of the Taylor seriess:
$$
f(\vec{a}+\vec{h})= f(\vec{a})+\int_0^1(1-t)\langle \vec{h},Hf(\vec{a}+t\vec{h})\vec{h} \rangle\,dt
$$

Oohh, that would have been really smart.

 

[graphking]

for question 10, I would like to mention that the hessian matrix is

It's also short if we use the integral remainder of the Taylor seriess:
$$
f(\vec{a}+\vec{h})= f(\vec{a})+\int_0^1(1-t)\langle \vec{h},Hf(\vec{a}+t\vec{h})\vec{h} \rangle\,dt
$$

I think peano or the intergral form of the remainder would just be as same efficient

 

[Office_Shredder]

If anyone feels intimidated by #8 because you're not sure what the definitions are, it only requires a small amount of linear algebra and calculus, see the spoiler for more.

Spoiler

$L^2([0,1])$ is the set of all functions $f(x)$ on $[0,1]$ such that $\int_0^1 |f(x)|^2 dx$ exists. This is a vector space, and has an inner product defined by $\left<f,g\right> = \int_0^1 f(x)\bar{g(x)}dx$. Then $1$ and $x$ are both in $L^2([0,1])$ and$\left<1,x\right>=\int_0^1 1 \times x dx =1/2$. So these vectors are not orthogonal.

 

[etotheipi]

probably wrong for #8 but I'll try anyway; by gram schmidt we can come up with an orthogonal basis $(v_1, v_2)$ for $K$ by setting $v_1 = 1$ and $v_2 = x - \langle x, 1 \rangle = x - 1/2$. then it's just
$$\pi^{\bot}(v) = \frac{\langle v, x - \frac{1}{2} \rangle}{|x - \frac{1}{2}|^2} (x - \frac{1}{2}) + \langle v , 1 \rangle$$then$$\begin{align*}
\int_0^1 (x-\frac{1}{2}) e^x dx &= \frac{3-e}{2} \\

\int_0^1 (x-\frac{1}{2})^2 dx &= \frac{1}{12} \\

\int_0^1 e^x dx = e-1
\end{align*}$$so you just get$$\pi^{\bot}(e^x) = 6(3-e)(x-\frac{1}{2}) + (e-1)$$lol idk if that's right, but it's 3am so cba to check atm

 

[fresh_42]

probably wrong for #8 but I'll try anyway; by gram schmidt we can come up with an orthogonal basis $(v_1, v_2)$ for $K$ by setting $v_1 = 1$ and $v_2 = x - \langle x, 1 \rangle = x - 1/2$. then it's just
$$\pi^{\bot}(v) = \frac{\langle v, x - \frac{1}{2} \rangle}{|x - \frac{1}{2}|^2} (x - \frac{1}{2}) + \langle v , 1 \rangle$$then$$\begin{align*}
\int_0^1 (x-\frac{1}{2}) e^x dx &= \frac{3-e}{2} \\

\int_0^1 (x-\frac{1}{2})^2 dx &= \frac{1}{12} \\

\int_0^1 e^x dx = e-1
\end{align*}$$so you just get$$\pi^{\bot}(e^x) = 6(3-e)(x-\frac{1}{2}) + (e-1)$$lol idk if that's right, but it's 3am so cba to check atm

It is correct. And here are the sorted results:
$$
\pi^\perp(v) = \langle v,1 \rangle 1 + 12 \;\langle v,x-\dfrac{1}{2}\rangle \left(x-\dfrac{1}{2}\right)
$$
$$
\pi^\perp(e^x)=6x(3-e) +4e-10
$$

 

[julian]

Another way of doing 7 b:

\begin{align*}
\| A x \|_2^2 &= x^T A^T A x
\end{align*}

Define $M = A^T A$. Obviously, $M^T = M$. As $M$ is real and symmetric there exist the matrix $R$ who's columns are eigenvectors of $M$ and such that $R^T R = \mathbb{1}$ and

\begin{align*}
M = R D R^T , \quad
D=
\begin{pmatrix}
\lambda_1 & & & \\
& \lambda_2 & & 0 \\
0 & & \ddots & \\
& & & \lambda_d
\end{pmatrix}
\end{align*}

where $\lambda_i$ are the eigenvalues. We make the following change of variables with unit Jacobian:

\begin{align*}
x' = R^T x \quad (\det R^T = 1)
\end{align*}

in

\begin{align*}
\int \prod_{i=1}^d dx_i \exp ( -x^T M x) &= \int \prod_{i=1}^d dx_i' \exp ( -x^{'T} D x')
\\
&= (\pi)^{d/2} / (\prod_{i=1}^d \lambda_i)^{1/2}
\\
&= (\pi)^{d/2} / (\det M)^{1/2}
\\
&= (\pi)^{d/2} / \det A .
\end{align*}

 

[benorin]

6. Let $f\in L^2 ( \mathbb{R} )$ and $g : \mathbb{R} \longrightarrow \overline{\mathbb{R}}$ be given as
$$
g(t):=t\int_\mathbb{R} \chi_{[t,\infty )}(|x|)\exp(-t^2(|x|+1))f(x)\,dx
$$
Show that $g\in L^1(\mathbb{R}).$

Spoiler: #6

Reference for this definition is Papa Rudin pg. 65:

Definition: If $0<p<\inf$ and if $f$ is a complex measurable function on the measure space $X$, define

$$\| f \|_{p}:=\left\{\int_{X} |f|^p\, d\mu\right\}^{\tfrac{1}{p}}$$

and let $L^p( \mu )$ consist of all $f$ for which $\| f\|_{p}<\infty$.

Work:
$$\begin{align} \| g\|_{1} & =\int_{\mathbb{R}}\left| t \int_{\mathbb{R}}\chi_{\left[ t,\infty \right)}(|x|) \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt \\ & =
\begin{cases}
\int_{\mathbb{R}}\left| t \left( \int_{-\infty}^{-t}+\int_{t}^{\infty}\right) \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt & \text{if } t \geq 0 \\
\int_{\mathbb{R}}\left| t \int_{-\infty}^{\infty} \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt & \text{if } t < 0
\end{cases} \\ & = 4\int_{0}^{\infty} t\left| \int_{\max\left\{ t,0 \right\} }^{\infty} \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt \\ & \leq 4\int_{0}^{\infty} t \int_{\max\left\{ t,0 \right\} }^{\infty} \exp (-t^2(|x|+1))\left| f(x)\right| \, dx \, dt \\ & \leq \underbrace{4\int_{0}^{\infty} t e^{-t^2}\, dt}_{=2}\cdot \int_{\max\left\{ t,0 \right\} }^{\infty} |f(x)|\, dx \\ & \leq 2\left\{\left[ \int_{\max\left\{ t,0 \right\} }^{\infty} |f(x)|^2\, dx\right]^{\tfrac{1}{2}}\right\} ^{2} < \infty \\ \end{align}$$

where the last inequality (finiteness) follow from the hypothesis that $f\in L^{2}(\mathbb{R} )$ and this was to be shown.

 

[fresh_42]

Spoiler: #6

Reference for this definition is Papa Rudin pg. 65:

Definition: If $0<p<\inf$ and if $f$ is a complex measurable function on the measure space $X$, define

$$\| f \|_{p}:=\left\{\int_{X} |f|^p\, d\mu\right\}^{\tfrac{1}{p}}$$

and let $L^p( \mu )$ consist of all $f$ for which $\| f\|_{p}<\infty$.

Work:
$$\begin{align} \| g\|_{1} & =\int_{\mathbb{R}}\left| t \int_{\mathbb{R}}\chi_{\left[ t,\infty \right)}(|x|) \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt \\ & =
\begin{cases}
\int_{\mathbb{R}}\left| t \left( \int_{-\infty}^{-t}+\int_{t}^{\infty}\right) \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt & \text{if } t \geq 0 \\
\int_{\mathbb{R}}\left| t \int_{-\infty}^{\infty} \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt & \text{if } t < 0
\end{cases} \\ & = 4\int_{0}^{\infty} t\left| \int_{\max\left\{ t,0 \right\} }^{\infty} \exp (-t^2(|x|+1)) f(x)\, dx \right| \, dt \\ & \leq 4\int_{0}^{\infty} t \int_{\max\left\{ t,0 \right\} }^{\infty} \exp (-t^2(|x|+1))\left| f(x)\right| \, dx \, dt \\ & \leq \underbrace{4\int_{0}^{\infty} t e^{-t^2}\, dt}_{=2}\cdot \int_{\max\left\{ t,0 \right\} }^{\infty} |f(x)|\, dx \\ & \leq 2\left\{\left[ \int_{\max\left\{ t,0 \right\} }^{\infty} |f(x)|^2\, dx\right]^{\tfrac{1}{2}}\right\} ^{2} < \infty \\ \end{align}$$

where the last inequality (finiteness) follow from the hypothesis that $f\in L^{2}(\mathbb{R} )$ and this was to be shown.

The last inequality's name is Hölder.

 

[Office_Shredder]

Don't the outermost square and square root cancel, and then it looks a lot to me like the second to last step is just asserting that
$$\int_{\max(t,0)}^\infty |f(x)| dx \leq \int_{\max(t,0)}^{\infty} |f(x)|^2 dx$$

I must be missing something because I don't think that's true.

 

[fresh_42]

Don't the outermost square and square root cancel, and then it looks a lot to me like the second to last step is just asserting that
$$\int_{\max(t,0)}^\infty |f(x)| dx \leq \int_{\max(t,0)}^{\infty} |f(x)|^2 dx$$

I must be missing something because I don't think that's true.

If we do not separate the factors, then we have Hölder ($1=1/2 +1/2$)
$$
\int_\mathbb{R} |u(x)f(x)|\,dx =\|uf\|_1\leq \|u\|_2\|f\|_2 < \infty
$$
where $u$ is the exponential factor.

 

[julian]

Another way of doing 7 b:

\begin{align*}
\| A x \|_2^2 &= x^T A^T A x
\end{align*}

Define $M = A^T A$. Obviously, $M^T = M$. As $M$ is real and symmetric there exist the matrix $R$ who's columns are eigenvectors of $M$ and such that $R^T R = \mathbb{1}$ and

\begin{align*}
M = R D R^T , \quad
D=
\begin{pmatrix}
\lambda_1 & & & \\
& \lambda_2 & & 0 \\
0 & & \ddots & \\
& & & \lambda_d
\end{pmatrix}
\end{align*}

where $\lambda_i$ are the eigenvalues. We make the following change of variables with unit Jacobian:

\begin{align*}
x' = R^T x \quad (\det R^T = 1)
\end{align*}

in

\begin{align*}
\int \prod_{i=1}^d dx_i \exp ( -x^T M x) &= \int \prod_{i=1}^d dx_i' \exp ( -x^{'T} D x')
\\
&= (\pi)^{d/2} / (\prod_{i=1}^d \lambda_i)^{1/2}
\\
&= (\pi)^{d/2} / (\det M)^{1/2}
\\
&= (\pi)^{d/2} / \det A .
\end{align*}

Here's my proof again. I should mention that in my proof I used Sylveter's criterion. In particular the theorem:

"A real-symmetric matrix $M$ has non-negative eigenvalues if and only if $M$ can be factored as $M = A^TA$, and all eigenvalues are positive if and only if $A$ is non-singular".

Also, I think the answer is supposed to be $\pi^{d/2} / |\det A|$.

 

[fresh_42]

I should mention that in my proof I used Sylveter's criterion. In particular the theorem:

"A real-symmetric matrix $M$ has non-negative eigenvalues if and only if $M$ can be factored as $M = A^TA$, and all eigenvalues are positive if and only if $A$ is non-singular".

Also, I think the answer is supoosed to be $\pi^{d/2} / |\det A|$.

The shortest version is probably by the transformation theorem for integrals. $\varphi (x)=Ax$ is $C^1$ and $D\varphi =A.$

 

[benorin]

If we do not separate the factors, then we have Hölder ($1=1/2 +1/2$)
$$
\int_\mathbb{R} |u(x)f(x)|\,dx =\|uf\|_1\leq \|u\|_2\|f\|_2 < \infty
$$
where $u$ is the exponential factor.

Look at you giving mana from the prof: I had no idea this was so useful here... ;)

@Office_Shredder Yes I indeed reasoned like you had and must have assumed $|f(x)|\geq 1$ in my head lol

 

[julian]

I should have just said, that in my proof of 7 b, I used: Given that $M = A^T A$, the eigenvalues of $M$ are non-negative because

\begin{align*}
\lambda = \frac{v^T M v}{v^T v} = \frac{v^T A^T A v}{v^T v} = \frac{\| A v \|_2^2}{\| v \|_2} \geq 0 .
\end{align*}

And the eigenvalues must be non-zero because $M$ is non-singular if $A$ is non-singular. That would have been more helpful to people. This is the reverse implication of the theorem:

"A real-symmetric matrix $M$ has non-negative eigenvalues if and only if $M$ can be factored as $M = A^TA$, and all eigenvalues are positive if and only if $A$ is non-singular"

And, at the very end of my calculation (post #38) I should have written down $\pi^{d/2} / |\det A|$ because $(\det M)^{1/2}$ is a positive number.

 

[julian]

The shortest version is probably by the transformation theorem for integrals. $\varphi (x)=Ax$ is $C^1$ and $D\varphi =A.$

So you are taking a vector valued function $\varphi (x)$ and considering the derivative of it. So in component form $(D \varphi)_{ij} = \partial_j \varphi_i (x) = A_{ij}$. Is that right? Where do you go from there?

 

[fresh_42]

So you are taking a vector valued function $\varphi (x)$ and considering the derivative of it. So in component form $(D \varphi)_{ij} = \partial_j \varphi_i (x) = A_{ij}$. Is that right? Where do you go from there?

I use
http://stat.math.uregina.ca/~kozdron/Teaching/Regina/351Fall15/Handouts/351lect05.pdf,
Tonelli, and $\|x\|_2^2=\sum x_i^2$. No need to go to the components of the Jacobi matrix.

 

[julian]

I use
http://stat.math.uregina.ca/~kozdron/Teaching/Regina/351Fall15/Handouts/351lect05.pdf,
Tonelli, and $\|x\|_2^2=\sum x_i^2$. No need to go to the components of the Jacobi matrix.

Oh, you are just talking about the method @graphking gave (post #16)? Yeah, yeah sure.

 

[Not anonymous]

13. Write

A doubt, possibly silly, about question 13. Are and required to be integers?

 

[fresh_42]

A doubt, possibly silly, about question 13. Are and required to be integers?

Well, it requires only inductions and quadratic equations, so it is technically not difficult. But it is quite tricky, I admit. 12, 14, and 15 are probably easier. 11 is easy with the fundamental theorem of algebra and determinants, so it is technically difficult, but not very tricky.

 

[Not anonymous]

Well, it requires only inductions and quadratic equations, so it is technically not difficult. But it is quite tricky, I admit. 12, 14, and 15 are probably easier. 11 is easy with the fundamental theorem of algebra and determinants, so it is technically difficult, but not very tricky.

Sorry again, my question was incomplete due to copy-paste quote error. Meant to ask whether $a, b, c, d$ in the solution for question 13 should be integers.

For question 11, are $a(x)$ and $b(x)$ too required to be polynomials, or just some real-valued functions? I am not familiar with the notation $\mathbb{R}[x]$. Searching on the web, I see that one forum defines $\mathbb{R}[x]$ as polynomial functions, but I would like to know if that is a standard convention used here in this math challenge too.

And thanks for those lightning quick replies!

 

[fresh_42]

Sorry again, my question was incomplete due to copy-paste quote error. Meant to ask whether $a, b, c, d$ in the solution for question 13 should be integers.

Yes. The problem would otherwise be trivial.

For question 11, are $a(x)$ and $b(x)$ too required to be polynomials, or just some real-valued functions? I am not familiar with the notation $\mathbb{R}[x]$. Searching on the web, I see that one forum defines $\mathbb{R}[x]$ as polynomial functions, but I would like to know if that is a standard convention used here in this math challenge too.

And thanks for those lightning quick replies!

Yes, $\mathbb{R}[x]$ stands for all polynomials with real coefficients.
$\mathbb{R}(x)$ would be quotients of polynomials with real coefficients, and $\mathbb{R}[[x]]$ are all formal power series with real coefficients, i.e. polynomials with an infinite degree.

 

[Not anonymous]

Spoiler: Question 13

Let $x = \sqrt[8] {2207 - \dfrac {1} {2207 - \dfrac {1} {2207 - \dfrac {1} {2207 - ...}}}}$. We note that $x^8 = 2207 - \dfrac {1} {2207 - \dfrac {1} {2207 - \dfrac {1} {2207 - ...}}}$ and this is equivalent to $x^8 = 2207 - \dfrac {1} {x^8} \Rightarrow x^8 + \dfrac {1} {x^8} = 2207$ (Equation 1).

Next, we note that $\left(y + \dfrac {1} {y}\right)^2 = y^2 + \dfrac {1} {y^2} + 2$. Applying this identity to equation 1 and using the fact that $x^8 = (x^4)^2$, we get $\left(x^4 + \dfrac {1} {x^4}\right)^2 = 2209 \Rightarrow \left(x^4 + \dfrac {1} {x^4}\right) = \pm 47$. Applying the same identity repeatedly, while considering only non-negative values for RHS (i.e. non-negative square root) in order to get real-valued solutions, we get $\left(x^2 + \dfrac {1} {x^2}\right) = \sqrt {47 + 2} = 7$, and therefore $\left(x + \dfrac {1} {x}\right) = \sqrt {7 + 2} = 3$. This is equivalent to the quadratic equation $x^2 - 3x + 1 = 0$, solving which we get $x = \dfrac {3 \pm \sqrt {5}} {2}$. Hence one solution of the form $\dfrac {a + b\sqrt{c}} {d}$ to the given $\sqrt[8] {2207 - ...}$ expression is with $a = 3, b = 1, c = 5, d = 2$

 

[Not anonymous]

Spoiler: Question 15

The positions of the first square in the 8 rows are $1,2,...,8$. For any given placement of 8 rooks such that none of those threaten another, the following 2 conditions are necessary and sufficient: (a) no two rooks must be in same row, (b) no two rooks must be in same column. Since there must be exactly one rook per row, we may denote the positions of 8 rooks by $pos_{i}$ where $i$ denotes the row and $pos_{i}$ is the number assigned to the square where the rook in that row is placed. Next we note that each position can be written as sum of number assigned to first square in that row and distance or offset of the rook from that first square, i.e. $pos_{i} = firstpos_{i} + offset{i}$. Here $firstpos_{i}$ would be $1, 2, .., 8$ respectively for the 8 rows in order and $offset_{i} \in \{0, 1, .., 7\}$. Now, since no two rooks can occupy the same column, we cannot have $offset_{i} = offset_{j}$ for any $i, j \in \{1, 2, ...,8\}, i \neq j$. Thus for any valid placement of 8 rooks, the offset values of the 8 rooks will be distinct and therefore since only 8 different possible values exist for offset values, they will cover the set $\{0, 1, 2, ...,7\}$, hence they will always sum to $0 + 1 + .. + 7 = 28$. Therefore, $S = \sum_{i=1}^8 pos_{i} = \sum_{i=1}^8 (firstpos_{i} + offset_{i}) =$

$\sum_{i=1}^8 firstpos_{i} + \sum_{i=1}^8 offset_{i} = (1 + 2 + .. + 8) + (0 + 1 + .. + 7) = 36 + 28 = 64$.

Any valid placement of rooks will always sum to this same value, 64.

 

[fresh_42]

Any valid placement of rooks will always sum to this same value, 64.

If I place one rook on $A1=1$ and another on $H8=64$ I already get $65$ with only $2$ rooks.

 

[etotheipi]

Think you just missed out a factor of 8 in your working, i.e. if $K$ is the set of positions $(i,j)$ with $i,j \in \{ 1,\dots, 8 \}$ then you ought to have $S = \sum_{(i,j) \in K} [i + 8(j-1)] = \sum_1^8 i + 8 \sum_1^7 j$

 

[Not anonymous]

If I place one rook on $A1=1$ and another on $H8=64$ I already get $65$ with only $2$ rooks.

Right, my answer was wrong. I was suspicious when I got the sum 64 as the answer since it seemed too small to be correct, but was half asleep and submitted the solution without cross-checking it by writing down on paper and summing up the example placement I had in mind. Below is a corrected version, hopefully correct this time.

Spoiler: Question 15

The positions of the first square in the 8 rows are $1,2,...,8$. For any given placement of 8 rooks such that none of those threaten another, the following 2 conditions are necessary and sufficient: (a) no two rooks must be in same row, (b) no two rooks must be in same column. Since there must be exactly one rook per row, we may denote the positions of 8 rooks by $pos_{i}$ where $i$ denotes the row and $pos_{i}$ is the number assigned to the square where the rook in that row is placed. Next we note that each position can be written as sum of number assigned to first square in that row and distance or offset of the rook from that first square, i.e. $pos_{i} = firstpos_{i} + offset_{i}$. Here $firstpos_{i}$ would be $\{1, 9, 17, 26, .., 57\}$ (numbers spaced apart by a difference of 8) respectively for the 8 rows in order and $offset_{i} \in \{0, 1, .., 7\}$. Now, since no two rooks can occupy the same column, we cannot have $offset_{i} = offset_{j}$ for any $i, j \in \{1, 2, ...,8\}, i \neq j$. Thus for any valid placement of 8 rooks, the offset values of the 8 rooks will be distinct and therefore since only 8 different possible values exist for offset values, they will cover the set $\{0, 1, 2, ...,7\}$, hence they will always sum to $0 + 1 + .. + 7 = 28$. Therefore, $S = \sum_{i=1}^8 pos_{i} = \sum_{i=1}^8 (firstpos_{i} + offset_{i}) =$

$\sum_{i=1}^8 firstpos_{i} + \sum_{i=1}^8 offset_{i} = (1 + 9 + 17 + .. + 57) + (0 + 1 + .. + 7) = 232 + 28 = 260$.

Any valid placement of rooks will always sum to this common value, 260.

Verified with the following two placements:
$\{A1, B2, C3, D4, E5, F6, G7, H8\} \rightarrow sum = (1 + 10 + 19 + 28 + 37 + 46 + 55 + 64) = 260$
$\{A1, B2, C3, D4, E5, F8, G6, H7\} \rightarrow sum = (1 + 10 + 19 + 28 + 37 + 48 + 54 + 63) = 260$

 

[Not anonymous]

Think you just missed out a factor of 8 in your working, i.e. if $K$ is the set of positions $(i,j)$ with $i,j \in \{ 1,\dots, 8 \}$ then you ought to have $S = \sum_{(i,j) \in K} [i + 8(j-1)] = \sum_1^8 i + 8 \sum_1^7 j$

Thanks! Yes, in the summation, I had mistakenly used as the start positions of rows the values $\{1, 2, .., 8\}$ instead of $\{1, 9, 17, ..., 57\}$, and that is equivalent to missing out a factor of 8 as you have mentioned

 

[mathwonk]

I thought I'd try again to drum up some interest in question #4. This subject is nice in that it mixes algebra and geometry and thus allows the two to enrich and illuminate each other. For example, this question is in the direction of showing the relation between factorization in algebra and decomposition in geometry. If you are at all interested, you might also try the slightly less abstract but more difficult problem that if f and g are polynomials in C[X,Y], where C is the complex numbers, and f is irreducible, then V(g) contains V(f) if and only if f divides g, (where V(f) is the set of zeroes of f in C^2, and V(g) is the set of zeroes of g.) In particular, if for all points p in the plane, f(p) = 0 implies also g(p) = 0, and f is irreducible, then f divides g. And give an example to show the irreducibility of f is needed for this statement. If f,g,...,h are distinct and irreducible, it follows that I(V(f^r.g^s...h^t)) = (f.g...h) = the principal ideal generated by the product of the distinct irreducible factors. Also V(f^r.g^s...h^t) = V(f) union V(g) ...union V(h), where each of V(f), V(g),...,V(h) is an irreducible variety.