it's okay despite not the most simple way :>Fycho wrote:
此虽然结果是证明X^3+Y^3<=2的，但是第二步已经证明出X^2+Y^2<=2....
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Topic Starter
Topic Starter
a more simple solution:
we have
x^2 + y^3 >= x^3 + y^4, which is
x^2 - x^3 >= y^4 - y^3 or y^3 - y^4 >= x^3 - x^2
also we have x >= 0, y >= 0 , that is
if x = 0 & y = 0 ==> x^2+y^2 = 0 < 2
if x > 0 & y = 0 ==> (x^2 >= x^3) ==> (x<1) ==> (x^2 + y^2 < 2)
same as y = 0 & x > 0
if x > 1 & y > 1 ==> inequality trivially does not hold
if x < 1 & y < 1 ==> inequality trivially hold
if x = 1, y^3 >= y^4 ==>
if y = 1, x^2 >= x^3 ==>
then
if x > 1 > y
1 - y > y - y^2 > y^2 - y^3 > y^3 - y^4 >= x^3 - x^2 > x^2 - x > x -1
we have
1 - y > x -1 ==> 2 > x+y
y - y^2 > x^2 - x ==> x+y > x^2 + y^2
put together, 2 > x^2 + y ^2
if
y > 1 > x
x^2 - x^3 >= y^4 - y^3
similar to above
we have
x^2 + y^3 >= x^3 + y^4, which is
x^2 - x^3 >= y^4 - y^3 or y^3 - y^4 >= x^3 - x^2
also we have x >= 0, y >= 0 , that is
if x = 0 & y = 0 ==> x^2+y^2 = 0 < 2
if x > 0 & y = 0 ==> (x^2 >= x^3) ==> (x<1) ==> (x^2 + y^2 < 2)
same as y = 0 & x > 0
if x > 1 & y > 1 ==> inequality trivially does not hold
if x < 1 & y < 1 ==> inequality trivially hold
if x = 1, y^3 >= y^4 ==>
if y = 1, x^2 >= x^3 ==>
then
if x > 1 > y
1 - y > y - y^2 > y^2 - y^3 > y^3 - y^4 >= x^3 - x^2 > x^2 - x > x -1
we have
1 - y > x -1 ==> 2 > x+y
y - y^2 > x^2 - x ==> x+y > x^2 + y^2
put together, 2 > x^2 + y ^2
if
y > 1 > x
x^2 - x^3 >= y^4 - y^3
similar to above
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
hope you'll like it.
if your queue is full,ignore me please.thanks in advance！
https://osu.ppy.sh/s/175720
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
It was a quizz, please read before.
Topic Starter
Two quiz for two modding tickets:
Rules:
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:
ps: x, y , z are integer
ps: x, y , z are integer
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:
if real solution exists, then
x^4+y^4=(x^2+y^2+sqr(2)xy)(x^2+y^2-sqr(2)xy)=z^2, show that the result of formula on the left is a perfect square
but (x^2+y^2+sqr(2)xy)≠(x^2+y^2-sqr(2)xy), and it can't be factorized anymore
so, there's no real solution
I haven't finished it yet .
One part is too difficult QAQ
Let me post it before I finished it all.
太纸张证不出来 edit了 让后面的人证把
One part is too difficult QAQ
太纸张证不出来 edit了 让后面的人证把
SPOILER
We suppose that the three reals(edit again here *rational numbers) are x=p/q y=r/s z=m/n
where (p,q)=1,(r,s)=1,(m,n)=1,p,q,r,s,m,n are integers;
then
x^4+y^4
=(p/q)^4+(r/s)^4
=p^4/q^4+r^4/s^4
=((ps)^4+(qr)^4)/(sq)^4
=((ps)^4+(qr)^4)/((sq)^2)^2
=z^2
So (ps)^4+(qr)^4=z^2*((sq)^2)^2
=(z*(sq)^2)^2
Since the left side are all integers, the right side should be an integer.
if x^2=t,t is an integer,we can easily know that x is an integer.
Thus , the question becomes that we have to find three integers a,b,c that satisfies
a^4+b^4=c^2
Obviously,a and b can't be two odd numbers.
If a is even and b is odd,then c is odd,
then a^4=c^2-b^4=(c-b^2)(c+b^2)
and We write a^4 as (2a0)^4=16*a0^4
We consider about mod 16.
Of course a^4==0 (mod 16) b^4==1(mod 16)
So c^2==1(mod 16) And that results in c==7(mod 8)
c=8c0-1 I can't prove this part ,sorry
If a and b are all even,
then we let a=2a0,b=2b0,c=2c0,
(2a0)^4+(2b0)^4=(2c0)^2
4a0^4+4b0^4=c0^2
and we know that c0 is even
So c0=2c1
We have a0^4+b0^4=c1^2 now, which has the same structure as a^4+b^4=c^2
In that occasion,we only have the solution a=b=c=0,but it don't satisfy with x>0,y>0.
where (p,q)=1,(r,s)=1,(m,n)=1,p,q,r,s,m,n are integers;
then
x^4+y^4
=(p/q)^4+(r/s)^4
=p^4/q^4+r^4/s^4
=((ps)^4+(qr)^4)/(sq)^4
=((ps)^4+(qr)^4)/((sq)^2)^2
=z^2
So (ps)^4+(qr)^4=z^2*((sq)^2)^2
=(z*(sq)^2)^2
Since the left side are all integers, the right side should be an integer.
if x^2=t,t is an integer,we can easily know that x is an integer.
Thus , the question becomes that we have to find three integers a,b,c that satisfies
a^4+b^4=c^2
Obviously,a and b can't be two odd numbers.
If a is even and b is odd,then c is odd,
and We write a^4 as (2a0)^4=16*a0^4
We consider about mod 16.
Of course a^4==0 (mod 16) b^4==1(mod 16)
So c^2==1(mod 16) And that results in c==7(mod 8)
c=8c0-1
If a and b are all even,
then we let a=2a0,b=2b0,c=2c0,
(2a0)^4+(2b0)^4=(2c0)^2
4a0^4+4b0^4=c0^2
and we know that c0 is even
So c0=2c1
We have a0^4+b0^4=c1^2 now, which has the same structure as a^4+b^4=c^2
In that occasion,we only have the solution a=b=c=0,but it don't satisfy with x>0,y>0.
卧槽
ps: x, y , z are integer
sco大大你怎么这么坑爹。。。OVQ
ps: x, y , z are integer
sco大大你怎么这么坑爹。。。OVQ
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
#1
1. とらドラ!
2. Clannad afterstory
3. 新世界より
/.\
1. とらドラ!
2. Clannad afterstory
3. 新世界より
/.\
QUIZ #1
1.tennis king son
2.ghost father
3.stone age
1.tennis king son
2.ghost father
3.stone age
不是要說明集數?
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
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
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
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
Topic Starter
yes still open but you need 30x100
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