why am i on the train when you post the easy games
Don't you even read. Which map is there above?pupil zone wrote:
in case you skipped some maps above...
hope you'll like it.
if your queue is full,ignore me please.thanks in advance!
https://osu.ppy.sh/s/175720
I hate mathScorpiour wrote:
Two quiz for two modding tickets:
Rules:
- You may not edit your post!
- There're two quiz in total, the one who posts the correct first will earn a modding ticket
- One answer per person, that is , you may not answer both questions , also means you may not both the tickets.
QUIZ #1Here are THREE screenshot from THREE different anime, post all the correct answer of ANIME Name with SEASON Number & EPISODE Number QUIZ #2Here's a mathematics quiz:
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