美女與野獸的故事。

老商人時運不濟，

所謂老就是年紀大呼吸過的空氣比放的屁還要多，

年輕林業生意做光山上只剩腐敗的紅檜，只剩不是鳥居的索道，

祖產沒了沒有錢了，只能巴望著女兒嫁給有錢人繼續安定過好日子。



野獸老爺不講道理拿著水壺便能四處小便，

看似沒有文化但有錢就是可以任性，身旁亦不缺美女但多個也無礙，

老商人女兒的美貌等於一大箱沈甸甸的人民幣。

美女盡職地嫁給野獸老爺，老商人的善良正義回來了，

沈默偉大歲月從先人墳墓慢慢流出，又是美妙時光，

空氣泛著輕飄的音樂，可能是擾亂空氣波的流行音樂吧。



最初，美女無法忍受野獸老爺的低水平，野獸！

但在百般順從愛情的影響下，野獸老爺終於變成韓系師奶殺手，

心由相生，開始大談抽象廢話追尋心靈價值，

再也沒有人莫名其妙的消失，

再也沒有人可以顛覆野獸苦蹲十年監牢。



這是古典童話故事，美女不可以選擇丈夫，女人必須充分發揮她的命運，

如果她不愛男人，無法順著神的劇本演人生大戲，婚姻家庭價值將會崩解，

信望盟會很生氣喔，不可以欺負多數人的沈默喔，科科。



空氣中突然嘔出腐敗的蘇格蘭威士忌味道，這是什麼西方思想？

沒有婚姻的美女，一文不值，

老商人把人民幣換成新台幣，

開心數錢，數好數滿，數到痔瘡都坐破了。

Notes on ellipse

1. Equation:

• x^2/a^2 + y^2/b^2 = 1
• x(t) = a cos(t), y(t) = b sin(t)

2. Arc length:

• Integrate from t = 0 to t_0, sqrt[x'(t)^2 + y'(t)^2] dt
• Integrate from t = 0 to t_0, sqrt[a^2 sin(t)^2 + b^2 cos(t)^2] dt

Fermat's theorem on sums of two squares

1. Introduction

• wiki: https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares
• A prime p is a sum of two integer squares if and only if p = 2 or p ≡ 1 (mod 4).

2. Brahmagupta–Fibonacci identity

• wiki: https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity
• A positive integer n is a sum of two squares if and only if ord_p(n) is even for all primes p ≡ 3 (mod 4).

3. Algorithm

• Reference: 
• http://www.ams.org/journals/mcom/1972-26-120/S0025-5718-1972-0314745-6/S0025-5718-1972-0314745-6.pdf
• http://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime/5883#5883
• Hermite-Serret Algorithm:
• Find x^2 ≡ -1 (mod p) where 0 < x < p/2.
• If a is a quadratic nonresidue modulo p then a^((p-1)/2) ≡ -1(mod p), so we can take x ≡ a^((p-1)/4) (mod p). 
• How does one find quadratic nonresidues? Well just over half of all numbers a such that 1< a < p-1 are quadratic nonresidues. 
• Taking a at random will need at most two goes on average.
• Carry out the Euclidean algorithm on p/x, producing the sequence of remainders R_1, R_2, ..., to the point where R_k is first less than Sqrt[p]. Then 
• p = (R_k)^2 + (R_{k+1})^2 if R_1 > 1, 
• p = x^2 + 1^2 if R_1 = 1.

4. Counting:

• Reference: 
• http://matwbn.icm.edu.pl/ksiazki/aa/aa40/aa4029.pdf
• http://mathworld.wolfram.com/SumofSquaresFunction.html

[2016-01-07] 北京政府操控Shanghai Stock Exchange Composite Index

如果突然有資金需求，錢被卡住真是叫天天不應。

Pythagorean triples

1. wiki:

• https://en.wikipedia.org/wiki/Pythagorean_triple
• https://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples

2. Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of positive integers m and n with m > n.

• a = m^2 - n^2
• b = 2mn
• c = m^2 + n^2

where m and n is coprime and odd parity.

How to iterate all coprime pairs:

• Farey sequence
• wiki: https://en.wikipedia.org/wiki/Farey_sequence
• Counting
• https://primes.utm.edu/notes/relprime.html
• 6 / pi^2

3. Counting:

• https://projecteuler.net/problem=540

算不出來的滋味。

沮喪的心情，往日你喜歡沈溺複雜公式，

成長的題數，曾百岳般刺激你的活力，

再不願讓妳征服！別害羞，快撤退，

你這老人每個地形架繩確保也要跌跌撞撞。



算了吧，我的心。

睡吧，呼呼鼾聲喊著喊著。



精神失敗，筋疲力盡！

這業餘數學家，成功再也無法引誘好勝的心，

台北的冬天失去霉鏽味的雨滴卻留住了寒冷，

寒冷呀，你可願意帶著我離開算不出來的滋味？

Notes on Möbius function

To solve ProjectEuler problems, we need some basic knowledge about Möbius function.

1. Definition: mu(n).



2. sum_{d|n} mu(d) = [n = 1].
Let n = gcd(a, b). sum_{d|gcd(a, b)} mu(d) = sum_{d|a, d|b} mu(d) = [gcd(a, b) = 1]



3. Mertens function: M(n) = sum_{k ≤ n} mu(k).



4. M(n) = 1 - sum_{d ≥ 2} M(n/d).
sum_{d ≤ n} M(n/d) = sum_{d ≤ n} sum_{k ≤ n/d} mu(k) = sum_{d ≤ n} sum_{k|d} mu(k) = sum_{d ≤ n} [d = 1] = 1

Reference: Deléglise, M., Rivat, J.: Computing the summation of the Möbius function. Experimental Mathematics 5(4), 291–295 (1996). [https://projecteuclid.org/download/pdf_1/euclid.em/1047565447]



5. Let S(n) denote the number of square-free positive integers ≤ n.
S(n) = sum_{d ≤ sqrt(n)} mu(d) floor(n/d^2).

Besides, let C(n) be the number of cube-free positive integers ≤ n.
C(n) = sum_{d ≤ n^{1/3}} mu(d) floor(n/d^3).

Reference: Jakub Pawlewicz, Counting square-free numbers (2011), arXiv:1107.4890. [http://arxiv.org/pdf/1107.4890v1.pdf]



6. Let P(n) be the number of primitive Pythagorean triples with a < b < c ≤ n.
Let a = x^2 − y^2, b = 2xy, c = x^2 + y^2 with x, y coprime (positive) integers of opposite parity.

Hence, P(n) = { (x, y) in R(n) : gcd(x, y) = 1, opposite parity } where R(n) = { (x, y) : x^2 + y^2 ≤ n, 0 < x < y }.
Also we let Q(n) = { (x, y) in R(n) : gcd(x, y) = 1 }.

If (x, y) in Q(n), then (1) x, y opposite parity (2) x, y are odd.  In case (1), (x, y) in P(n).  In case (2), gcd(a, b, c) = gcd(x^2 - y^2, 2xy, x^2 + y^2) = gcd(2x^2, 2xy, 2y^2) = 2 gcd(x^2, xy, y^2) = 2.  So (x, y) in P(n/2).  Hence Q(n) = P(n) + P(n/2).

Therefore, P(n) = sum_{k ≥ 0}(-1)^k Q(n/2^k). (no more opposite parity).

Now we want to represent Q(n) in terms of R(n): Q(n) = sum_{1 ≤ k ≤ Floor[Sqrt[n]]} mu(k) R(n/k^2).

Reference: Pythagorean triangles with legs less than n [http://ac.els-cdn.com/S0377042701004964/1-s2.0-S0377042701004964-main.pdf?_tid=2a2d0446-ac6b-11e5-a15e-00000aab0f27&acdnat=1451201377_754b996c3258324dc6694d8b83ba6b6f]