Solution for Problem #2

For those of you guessing 30, yes your answer is correct.

Proof of minimality

Let $S$ be the set of players in the group, but we will start from the empty set ($S = \emptyset$) and add new players as we advance from week to week. For the first month, we need to add $20$ different individuals to cover all $5 \cdot 4$ slots since everyone can only attend once every month. Now for the second month, we will try to analyze week per week. Let $S_n$ be the set of people (already in the set $S$) that is available to be added into $S$ at the start of week $n$  :

- Week 1, $S_1 = \{A_1,A_2,A_3,A_4,A_5\}$. We can only add one $A_i$ since they already met at the week 1 of the 1st month, (any of them will be the same, so let's add $A_1$). Then we need to add $4$ extra guys, let's call them $E_1,E_2,E_3,E_4$

- Week 2, $S_2 = \{A_2,A_3,A_4,A_5,B_1,B_2,B_3,B_4,B_5\}$. By similar fashion, we can only add one $A_i$, and one $B_j$ from $S_2$. Without losing of generality, let's add $A_2$ and $B_1$, then add $3$ new guys $F_1,F_2,F_3$

Now do you see the pattern ?
For Week 3, we will just add $A_3,B_2,C_1$, then add $2$ new guys $G_1,G_2$.
And for Week 4, add $A_4,B_3,C_2,D_1$ and add one new guy $H_1$.

We have $30$ people in $S$ after completing the 2nd month. So, we can claim that the number of people in the group should at least be $30$.

Now a possible configuration is the following :

1st Month

$A_1 A_2 A_3 A_4 A_5$

$B_1 B_2 B_3 B_4 B_5$

$C_1 C_2 C_3 C_4 C_5$

$D_1 D_2 D_3 D_4 D_5$

2nd Month

$A_1 E_1 E_2 E_3 E_4$

$A_2 B_1 F_1 F_2 F_3$

$A_3 B_2 C_1 G_1 G_2$

$A_4 B_3 C_2 D_1 H_1$

3rd Month

$A_5 B_4 C_3 D_2 E_1$

$B_5 C_4 D_3 E_2 F_2$

$C_5 D_4 E_3 F_3 G_1$

$D_5 E_4 F_4 G_2 H_1$

Note that we can modify the arrangement for the 3rd month, such that we don't need to add more people to $S$ after the 2nd month, and yet it will still meet the requirements on the problem. One important point to help you is that, since the configuration for the 4th month will be exactly the same with that of 1st month, then we have some 'restrictions' for the weeks on the 3rd month :

-  $A_i$ can attend at most until the 1st week

-  $B_i$ can attend at most until the 2nd week

-  $C_i$ can attend at most until the 3rd week

-  $D_i$ can attend at most until the 4th week

So for example, since $A_i$ can only attend the 1st week of the 3rd month, we should use the only available $A_i$ at that week i.e. $A_5$. Applying this idea and by tracking all $S_n$ as in explained in the 2nd month, will automatically generate the above's configuration for you. 

Problem #2 - How big is this Group ?

Don't worry, it's nothing about Group Theory. The following is an interesting combinatorics problem that I got few moments ago from one of my friend, Leonardo

"If every week $5$ person from a group are to meet with each other and they are assigned such that a person $x$ need to attend the meeting at most once a month and he/she will not see the same people ( either of the people) from previous meeting for at least $3$ months, what is the minimum number of the people in such group?" (1 month = 4 weeks)

Some clarification :

- If $A$ attends a meeting at week $W$, he can only attend again at least at week $W+4$ (so not only once a month)

- If both $A$ and $B$ meet at a meeting in week $W$, they can only meet again at week $W+13$

One simple fact is the group must have 20 distinct people to be assigned into 4 weeks of the first month (5 people each). The challenge is to find the minimum number of people in such group. Equivalently, if that minimum number is $N$, there will be no configurations satisfying the requirements in the problem with $N-1$ people.

See Solution

Problem #1 - Tackling the Diophantine

"Mathematics is the queen of the sciences and number theory is the queen of mathematics." - G.H. Hardy

Even though I might not concentrate in Number Theory during my upcoming PhD years, but what can I say, she is still my 'first love'. So let's get started with a Diophantine Equation !

I found this problem in my latest homework about Ring Theory especially interesting :

Find the integer solutions (x,y) of

\[x^3-2 = y^2\]

There is an advanced solution revolving around the ring $ \mathbb{Z}[\sqrt{-2}] $, which I just found recently (you don't wanna hear where I solved it), but the challenge is can you give a solution using Elementary Number Theory ?

For your information, there are exactly 2 pairs of integer solutions for this equation.

You're up !

Welcome to Math District 91

"Parts of Math can be simple because either triviality is normal, or everything is normal. But if you are just rational, you need to be real and much more, because Math is complex. Math is not always about numbers, but you can say that, up to isomorphism. Believe it or not, you can't see Math from only the right angle because Math is infinitely generated. However, enumerating combinations of open problems can be a base to help you unlocking a new dimension in Math." - Raymond Christopher

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