Mathematical Girls - Fermat's Last Theorem

Is this series still ongoing and dropped?

Yes, but can you prove that assertion (successfully)?

Only three fields for the groups, so I can only give the other two groups a mention here in the comments, they are Forgotten Scans and MangaIchi Scanlation Division.

2edgy4me.
QED.

Yes, but can you prove that assertion (successfully)?

2edgy4me.
QED.

I'm... I'm too stupid for this manga

Nope, you will be stupid only if you don't try to understand it.

 Miruka-san is best waifu. 

*looks at manga* *looks at comments* *looks back manga* We've been successfully invaded by mathematicians.

Yes, but can you prove that assertion (successfully)?

Yo, what is the answer on chapter 1 page 15 ? was just wondering since he didn't reveal it

I find these kind of puzzles kind of pointless, since it's often possible to find reasons for more than one of the options to be an 'odd one out'.

After all, what exactly stops me from defining a set A to contain all but one of the options, and then declare the final option the 'odd one out'?

Or I could take the polynomial y = (x-239)(x-251)(x-257)(x-263)(x-283), and declare 271 the 'odd one out' as the only one which is not a zero of my polynomial.

 

But I digress. I took a look at these numbers, and determined (what I think is) the most promising solution is 257 is the odd one out, as it is the only one not of the form 4n+3, where n is an integer. Still, this is slightly unsatisfying as it does not make mention of the fact that all these integers are primes.

On the other hand, 283 is the only one not of the form 3n + 200.  Ambiguity again!

But that's ok; you're not really supposed to find "the" right answer to these.  Coming up with a bunch of unique properties for each number in the list is a more productive thought exercise anyway.

Yo, what is the answer on chapter 1 page 15 ? was just wondering since he didn't reveal it

I find these kind of puzzles kind of pointless, since it's often possible to find reasons for more than one of the options to be an 'odd one out'.

After all, what exactly stops me from defining a set A to contain all but one of the options, and then declare the final option the 'odd one out'?

Or I could take the polynomial y = (x-239)(x-251)(x-257)(x-263)(x-283), and declare 271 the 'odd one out' as the only one which is not a zero of my polynomial.

But I digress. I took a look at these numbers, and determined (what I think is) the most promising solution is 257 is the odd one out, as it is the only one not of the form 4n+3, where n is an integer. Still, this is slightly unsatisfying as it does not make mention of the fact that all these integers are primes.

Yo, what is the answer on chapter 1 page 15 ? was just wondering since he didn't reveal it

I cannot say that i like Tetra's new character design...
Also aren't basic things like GCD (greatest common divisor) and LCM (least common multiple) tought in elementary school or something? I mean, they appear naturally when you add fractions...
That sneaky teacher...

Spoiler

It's actually the same problem...

I'm... I'm too stupid for this manga

Spoiler

I'm thinking of the difference between two adjacent square number (I don't know if that's a correct term). 

 

2^2 - 1^2 = 3

3^2 - 2^2 = 5

4^2 - 3^2 = 7

5^2 - 4^2 = 9 = 3^2

6^2 - 5^2 = 11

7^2 - 6^2 = 13

8^2 - 7^2 = 15

9^2 - 8^2 = 17

10^2 - 9^2 = 19

11^2 - 10^2 = 21

12^2 - 11^2 = 23

13^2 - 12^2 = 25 = 5^2

etc.

Since the difference contains every odd number, except for 1, there should be an infinite pythagorean triples as there are infinite odd square number. Of course that doesn't include all pythagorean triples such as 8,15,17.

I'm not a math major so excuse me for my confusing explanation.

Nah, you've got the right idea, just reworded in a different way. Considering two of the values a,b,c are consecutive, they will be coprime for sure. And from there, it's not a far leap to show that the difference will be coprime too.

Spoiler

It's weird being a math major and reading this. 

 

Any way, for the first question for this chapter, there are infinite primitive pythagorean triples. Think about it this way:

 

1 = 1^2

1 + 3 = 2^2

1 + 3 + 5 = 3^2

1 + 3 + 5 + 7 = 4^2

1 + 3 + 5 + 7 + 9 = 5^2

 

And so on and so forth (which can be proved via induction). Therefore, if the last term in the sum is a perfect square, we can write our sum now as 

 

(1 + 3 + 5 + 7) + 9 = 4^2 + 3^2 = 5^2, and we have our first primitive pythagorean triple. Since this pattern is infinite, as there are infinitely many odd perfect squares, there will always be infinite primitive triples as well. There are more primitive triples that can be found without using this method, but this is a surefire method.

 

The second problem of the chapter ties into this problem. We can divide the Pythagorean equation a^2 + b^2 = c^2 by c^2 to get (a/c)^2 + (b/c)^2 = 1, where (a/c, b/c) is a rational point on the unit circle. Aaaaand since we just showed that there are infinite primitive triples, there will be infinite rational points as well using this method.

 

I understand that this isn't a true "valid" proof, as we haven't explicitly shown that the sum of odd numbers method gives us coprime values every time, but this is a website where we have people legitimately following Black Clover and reading To Love Ru Darkness. I don't think we have to explain further.

 

Anyway, apologies in advance if chapter 3 focuses on the answers for these and I spoiled you all.

I'm thinking of the difference between two adjacent square number (I don't know if that's a correct term). 

Since the difference contains every odd number, except for 1, there should be an infinite pythagorean triples as there are infinite odd square number. Of course that doesn't include all pythagorean triples such as 8,15,17.

I'm not a math major so excuse me for my confusing explanation.

I can't wait until we get to abstract algebra

tbh im not sure how he didnt immediately figure out the answer to the unit circle problem despite JUST DISCUSSING primitive Pythagorean triples

It's weird being a math major and reading this. 

Any way, for the first question for this chapter, there are infinite primitive pythagorean triples. Think about it this way:

1 = 1^2

1 + 3 = 2^2

1 + 3 + 5 = 3^2

1 + 3 + 5 + 7 = 4^2

1 + 3 + 5 + 7 + 9 = 5^2

And so on and so forth (which can be proved via induction). Therefore, if the last term in the sum is a perfect square, we can write our sum now as 

(1 + 3 + 5 + 7) + 9 = 4^2 + 3^2 = 5^2, and we have our first primitive pythagorean triple. Since this pattern is infinite, as there are infinitely many odd perfect squares, there will always be infinite primitive triples as well. There are more primitive triples that can be found without using this method, but this is a surefire method.

The second problem of the chapter ties into this problem. We can divide the Pythagorean equation a^2 + b^2 = c^2 by c^2 to get (a/c)^2 + (b/c)^2 = 1, where (a/c, b/c) is a rational point on the unit circle. Aaaaand since we just showed that there are infinite primitive triples, there will be infinite rational points as well using this method.

I understand that this isn't a true "valid" proof, as we haven't explicitly shown that the sum of odd numbers method gives us coprime values every time, but this is a website where we have people legitimately following Black Clover and reading To Love Ru Darkness. I don't think we have to explain further.

Anyway, apologies in advance if chapter 3 focuses on the answers for these and I spoiled you all.

Not surprised it's only three volumes. Forget most kids not liking to use their brains, even most adults would find no interest in this. This is an amazingly small niche work.

they're just trying to teach me math. there's no story here.

I wonder what did you expect from a manga with this title.