# Mathematical Girls - Fermat's Last Theorem

Alt Names: | Suugaku Girl - Fermat no Saishuu Teiri 数学ガール フェルマーの最終定理 |

Author: | YUUKI Hiroshi |

Artist: | KASUGA Shun |

Genres: | Harem School Life Seinen |

Type: | Manga (Japanese) |

Status: | Ongoing |

Description: | "I" (Boku) love math. Just after the high school entrance ceremony, "I" meet a beautiful girl: Milka. Milka is a mathematical genius. She gives me many maths problems, she shows me many elegant solutions. Milka and I spend a long time by discussing maths in the school library. One year later, I meet another mathematical girl: Tetra. Tetra is one year younger than me, and asks me to teach her mathematics. While I teach her, she begins to understand mathematics and gradually to love its elegance. In this second volume (series), we talk about Pythagoras' Theorem, Elementary Number Theory, Group, Ring, Field, and Fermat's Last Theorem. |

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## 44 Comments

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.

I would say that your comment is far more "2edgy4me" than his which, frankly, wasn't at all.

2edgy4me.

QED.

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

Miruka-san is best waifu.

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

He reveal the answer on page 33 of the same chapter.

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.

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

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.

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.

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

Milka, huh? I bet she’s ... sweet.

*badum tish*xD

Alright, I’ll leave now.