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Scorpiour's Modding Queue (Modding Team)
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Topic Starter
For the Anime...
First one's Toradora. It has only 1 Season. It's Episode 7.
Second one AnoHana. Also just 1 Season. Episode 11.
Third is... Is...
Black Lagoon Ep. 7??
First one's Toradora. It has only 1 Season. It's Episode 7.
Second one AnoHana. Also just 1 Season. Episode 11.
Third is... Is...
Black Lagoon Ep. 7??
Topic Starter
OniJAM wrote:
If there are 2 coprime integers x,y satisfied (x^2)^2+(y^2)^2=z^2
so x^2=a^2-b^2, y^2=2ab, z=a^2+b^2 (a>b>0; a, b are coprime integers)
y must be even number; a, b have different parity
from x^2=a^2-b^2 we can see a must be odd and b must be even
also from here we know x^2+b^2=a^2
so x^2=m^2-n^2, b^2=2mn, a=m^2+n^2 (m>n>0; m, n are coprime integers)
so y^2=2ab=4mn(m^2+n^2)
so m, n, (m^2+n^2) are perfect squares
suppose m=p^2; n=q^2; m^2+n^2=r^2. (p,q,r) also satsfied x^4+y^4=z^2
but z=a^2+b^2=(m^2+n^2)^2+4m^2n^2>r^4>r>0
so from z we can always find a smaller integer satisfied the equation
so, there's no solution
kamisamaaa wrote:
数学题假设 (x,y,z)为方程(x^2)^2+(y^2)^2=z^2一个解并且x,y互质,y为偶数，则
x^2=a^2-b^2;y^2=2ab;z=a^2+b^2，其中a>b>0,a，b互质，a、b 的奇偶性相反。
由x^2=a^2-b^2得a必定是奇数，b必定是偶数。
另外，亦得x^2+ b^2=a^2，再从此得x=c^2-d^2;b=2cd;a=c^2+d^2，其中c>d>0，c,d互质，c、d的奇偶性相反。
因而y^2=2ab=4cd(c^2 + d^2)，
由此得c、d和c^2+d^2为平方数。
于是可设c=e^2;d=f^2;c^2+d^2=g^2,即e^4+f^4=g^2。
换句话说，(e,f,g)为方程x^4+^y^4=z^2的另外一个解。
但是，z=a^2+b^2=(c^2+d^2)^2+4c^2d^2>g^4>g>0。
就是说如果我们从一个z值出发，必定可以找到一个更小的数值 g,使它仍然满足方程x^4+y^4=z^2。如此类推，我们可以找到一个比g更小的数值，同时满足上式。
但是，这是不可能的！因为z为一有限值，这个数值不能无穷地递降下去 由此可知我们最初的假设不正确。所以，方程x^4+y^4=z^2没有正整数解
Prove not complete: you didn't explain why a & b are coprime and m &n are coprime either.
I wonder if Sco is an S while I'm an M.
Edit: No wonder, we are.
Edit: No wonder, we are.
Topic Starter
you may not post multiple times and your answer was googled :>OniJAM wrote:
odd^2 mod 4 == 1 <- (2t+1)^2=4t^2+4t+1 mod 4 == 1
even^2 mod 4 == 0 <- (2t)^2=4t^2 mod 4 == 0 (t>=0 and t is integer)
so x,y,z are all evens, or z and one of x,y are odds
if x,y are not coprime integers, we can extract the common divisor because z must have the same common divisor
so we can suppose x is odd, y is even, they are coprime integers
m&n same as above
哎，居然已经有人做出来了啊噗
其实他们证的都是正确的啦，虽然只是证明了x,y,z互质的情况下233，不过其实是一样的噗
（我自己做的时候绕在这里绕了半天才绕出来，结果一看他们的答案发现我是反着推上去的，然后就懵了噗。。。
设(x^2)^2+(y^2)^2=z^2存在正整数解
构建勾股数：
x^2=a^2-b^2
y^2=2ab
z =a^2+b^2
设a,b最大公因数为t，a=tm, b=tn, m,n互质
x^2=t^2*(m^2-n^2)
y^2=t^2*(2mn)
z =t^2*(m^2+n^2)
又由z=sqr((x^2)^2+(y^2)^2)=t^2*sqr((m^2-n^2)^2+(2mn)^2)存在正整数解，得z存在因数t^2
则设x/t=p, y/t=q, z/t^2=r，p,q,r互质
有
p^2=m^2-n^2
q^2=2mn
r =m^2+n^2
（接下来的过程我和他们的就一样啦，反正就是勾股数化为勾股素数，然后再往小逼近发现循环了于是就是假设错了，这么个意思，我就不写了
超讨厌数论啊噗
sco大大又要用超简单的证明过程来鄙视我们了噗
赖皮法：由费马大定理得x^4+y^4=a^4无正整数解，设x,y为正整数，则a无正整数解，即a^2=z无正整数解
其实他们证的都是正确的啦，虽然只是证明了x,y,z互质的情况下233，不过其实是一样的噗
（我自己做的时候绕在这里绕了半天才绕出来，结果一看他们的答案发现我是反着推上去的，然后就懵了噗。。。
设(x^2)^2+(y^2)^2=z^2存在正整数解
构建勾股数：
x^2=a^2-b^2
y^2=2ab
z =a^2+b^2
设a,b最大公因数为t，a=tm, b=tn, m,n互质
x^2=t^2*(m^2-n^2)
y^2=t^2*(2mn)
z =t^2*(m^2+n^2)
又由z=sqr((x^2)^2+(y^2)^2)=t^2*sqr((m^2-n^2)^2+(2mn)^2)存在正整数解，得z存在因数t^2
则设x/t=p, y/t=q, z/t^2=r，p,q,r互质
有
p^2=m^2-n^2
q^2=2mn
r =m^2+n^2
（接下来的过程我和他们的就一样啦，反正就是勾股数化为勾股素数，然后再往小逼近发现循环了于是就是假设错了，这么个意思，我就不写了
超讨厌数论啊噗
sco大大又要用超简单的证明过程来鄙视我们了噗
赖皮法：由费马大定理得x^4+y^4=a^4无正整数解，设x,y为正整数，则a无正整数解，即a^2=z无正整数解
blahblahblah, just to make things clear: x^2+y^2=z^2 is a pythagorean triple. in geometry there are not so many combbinations of coprime numbers for pythagorean triple, rest is said couple of times before.
solution to game #2
SPOILER
Let's assume that (x,y,z) with x, y, z > 0 is the integer solution to the equation that MINIMIZES z then:
1. if GCD(x,y) = d != 1, then (x/d , y/d , z/d^2) is still an integer solution to the equation.
so GCD(x,y) = 1
2. (x^2, y^2, z) is a pythagorean triple, hence they exist a,b INTEGER with 0<b<a that satisfy x^2 = 2ab , y^2 = a^2 - b^2 , z = a^2 + b^2, with GCD(a,b) = 1 and a !≡ b (mod 2) (if a is even then b is odd and vice versa). If a is even, then b is odd, so y^2 ≡ a^2 - b^2 ≡ -1 ≡ 3 (mod 4) which is an absurd because a square is always ≡ 0 or 1 (mod 4)
so they exist a,b INTEGER with 0<b<a that satisfy x^2 = 2ab , y^2 = a^2 - b^2 , z = a^2 + b^2, with GCD(a,b) = 1, with a odd and b even
3. as seen on 2, (y,b,a) is a pythagorean triple
so they exist u, v INTEGER with 0<v<u that satisfy b = 2uv , y = u^2 - v^2, a = u^2 + v^2, with GCD(u,v) = 1, with u !≡ v (mod 2)
From 2 and 3, we have that
x^2 = 4uv(u^2 + v^2)
since GCD(u,v) = 1, GCD(u, u^2+v^2) = GCD(v, u^2+v^2) = 1 -> so u, v and u^2+v^2 are squares of integer numbers
u = p^2 , v = q^2 , u^2+v^2 = r^2
from this, we have that p^4 + q^4 = r^2
r <= r^2 = u^2+v^2 = a <= a^2 < a^2 + b^2 = z
we found a smaller solution, which contraddicts the underlined hypothesis in the first sentence.
QED
1. if GCD(x,y) = d != 1, then (x/d , y/d , z/d^2) is still an integer solution to the equation.
so GCD(x,y) = 1
2. (x^2, y^2, z) is a pythagorean triple, hence they exist a,b INTEGER with 0<b<a that satisfy x^2 = 2ab , y^2 = a^2 - b^2 , z = a^2 + b^2, with GCD(a,b) = 1 and a !≡ b (mod 2) (if a is even then b is odd and vice versa). If a is even, then b is odd, so y^2 ≡ a^2 - b^2 ≡ -1 ≡ 3 (mod 4) which is an absurd because a square is always ≡ 0 or 1 (mod 4)
so they exist a,b INTEGER with 0<b<a that satisfy x^2 = 2ab , y^2 = a^2 - b^2 , z = a^2 + b^2, with GCD(a,b) = 1, with a odd and b even
3. as seen on 2, (y,b,a) is a pythagorean triple
so they exist u, v INTEGER with 0<v<u that satisfy b = 2uv , y = u^2 - v^2, a = u^2 + v^2, with GCD(u,v) = 1, with u !≡ v (mod 2)
From 2 and 3, we have that
x^2 = 4uv(u^2 + v^2)
since GCD(u,v) = 1, GCD(u, u^2+v^2) = GCD(v, u^2+v^2) = 1 -> so u, v and u^2+v^2 are squares of integer numbers
u = p^2 , v = q^2 , u^2+v^2 = r^2
from this, we have that p^4 + q^4 = r^2
r <= r^2 = u^2+v^2 = a <= a^2 < a^2 + b^2 = z
we found a smaller solution, which contraddicts the underlined hypothesis in the first sentence.
QED
Topic Starter
Yes these solutions are correct however most of the solution you can find online omits an important step that prove why x & y are co-prime which i expected :>
I'm not sure why you think proving x and y are coprimes is so important, it's trivial in that if they are not, you can reduce the whole equation until they are.Scorpiour wrote:
Yes these solutions are correct however most of the solution you can find online omits an important step that prove why x & y are co-prime which i expected :>
Also the solutions would be easily found online because your problem is a popular one which is a few hundred years old..
if they aren't coprime the triple that you chose in the beginning is not the one that minimizes z, so it's an absurd
Topic Starter
Your second line explained why i emphasized the coprimes proving ~KRZY wrote:
I'm not sure why you think proving x and y are coprimes is so important, it's trivial in that if they are not, you can reduce the whole equation until they are.Scorpiour wrote:
Yes these solutions are correct however most of the solution you can find online omits an important step that prove why x & y are co-prime which i expected :>
Also the solutions would be easily found online because your problem is a popular one which is a few hundred years old..
SPOILER
from my previous solution
you can write the bolded part thanks to the fact that u and v are coprime
to ensure that, you need a primitive pythagorean triple at point 3, and to ensure this you need a primitive pythagorean triple at point 2.
that pythagorean triple is primitive if and only if x and y are coprime
you can write the bolded part thanks to the fact that u and v are coprime
to ensure that, you need a primitive pythagorean triple at point 3, and to ensure this you need a primitive pythagorean triple at point 2.
that pythagorean triple is primitive if and only if x and y are coprime
Topic Starter
That prove is what i'm looking for but i never expect you guys try the answer on math quiz or find quiz once and once again o_o
watScorpiour wrote:
That prove is what i'm looking for but i never expect you guys try the answer on math quiz or find quiz once and once again o_o
Topic Starter
somebody argues me that "Scorpiour is just use the mappers for his amusement".
well, i will review the answers of my last math quiz again to decide who may win the ticket, and the anime puzzle ticket is still available
however there's no more quiz here anymore.
GG
well, i will review the answers of my last math quiz again to decide who may win the ticket, and the anime puzzle ticket is still available
however there's no more quiz here anymore.
GG
I think just because the quizzes are too difficult and really cost mappers' time.Scorpiour wrote:
somebody argues me that "Scorpiour is just use the mappers for his amusement".
GG
Answering a question is really fun,but answering a rather difficult question seems not so cool.
Some people spend half a day and even one day on your quizzes but can't find the answers,of course they will not feel happy.
I think you should make the quizzes easier,or make the answers Open-ended, then everyone will be willing to answer them and won't worry about whether they answer right or wrong.
Well,in my opinion,I like the way you giving tickets,if you can make those quizzes easier.
I really liked the (root2 + root3)^100 problem, although I didn't try to solve it.
1.とらドラ! Season 1- episode7
2.夏目友人帐 Season 1- episode7
3.薄樱鬼 Season 1- episode7
2.夏目友人帐 Season 1- episode7
3.薄樱鬼 Season 1- episode7
Topic Starter
incorrect
Topic Starter
/me sighs
somebody issues me closed the quiz queue.. what can i do
somebody issues me closed the quiz queue.. what can i do
Topic Starter
well , i decide to ignore others comment.
I never force anyone to participate so
more quiz with modding tickets are coming
I never force anyone to participate so
more quiz with modding tickets are coming
好
Topic Starter
i'm still evaluating your answerSephibro wrote:
so nobody won the number theory round?
伊原摩耶花 冰果
秋山澪 KON
金闪闪 fz
秋山澪 KON
金闪闪 fz
1.冰果 伊原摩耶花
2.夏色奇跡 花木优香
3.魔法禁書目錄 土御门元春
2.夏色奇跡 花木优香
3.魔法禁書目錄 土御门元春
Ibara Mayaka - Hyouka
Akiyama Mio - K-ON
Tsuchimikado Motoharu - ToAru no Index
Akiyama Mio - K-ON
Tsuchimikado Motoharu - ToAru no Index
Topic Starter
goodErunamoJAZZ wrote:
Itryhttp://osu.ppy.sh/s/107979
1.ice cream
2.hehe
3.coin
2.hehe
3.coin