QUIZ#1
1. 龙与虎
2. 夏目友人帐
3. 僵尸哪有那么萌?
看着像 乱书的 0.0
1. 龙与虎
2. 夏目友人帐
3. 僵尸哪有那么萌?
看着像 乱书的 0.0
You guys obviously didn't read this well, you are supposed to also give the episode number and the season number.Scorpiour wrote:
Here are THREE screenshot from THREE different anime, post all the correct answer of ANIME Name with SEASON Number & EPISODE Number
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
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
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 :>
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..
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
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