讓我看看....
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由 hycheah2000 於 星期一 十一月 29, 2004 2:09 am
由 訪客 於 星期二 十一月 09, 2004 9:53 pm
0,0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
由 lovebetty 於 星期一 十一月 08, 2004 10:51 pm
SEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
由 10607 於 星期三 八月 11, 2004 5:32 pm
無限多解嗎....不知道押.....看看
[其它]see
由 小吉爾 於 星期三 十二月 03, 2003 6:20 pm
see!!
由 lesterlau 於 星期日 十一月 30, 2003 3:33 pm
......
由 傻佬 於 星期日 十一月 16, 2003 12:10 pm
cc
由 紅樓綠園 於 星期日 十一月 16, 2003 10:38 am
as c
由 神乎其技 於 星期六 六月 07, 2003 6:24 pm
see
由 ---- 於 星期二 五月 27, 2003 9:44 pm
That's just an estimation~
由 ---- 於 星期二 五月 27, 2003 9:43 pm
Then, how many integer solutions for 3M^2+2N^2=350000 ??
2N^2 is even, 350000 is even. So 3M^2 is even.
Let M=2a
12a^2+2N^2=350000
6a^2+N^2=175000
6a^2 is even, 175000 is even. So N^2 is even.
Let N=2b
6a^2+4b^2=175000
3a^2+2b^2=87500
2b^2 is even. 87500 is even. So 3a^2 is even.
Let a=2c
12c^2+2b^2=87500
6c^2+b^2=43750
6c^2 is even. 43750 is even. So b^2 is even.
Let b=2d
6c^2+4d^2=43750
3c^2+2d^2=21875
4c=M, 4d=N
If c =3k
3c^2=0(mod 27)
21875=5(mod 27)
So 2d^2=5(mod 27)
2d^2<21875
2d^2=27e+5
e is odd.
e<810
Let e=2f+1
2d^2=54f+32
d^2=27f+16
2f+1<810
f<=404
(d-4)(d+4)=27f
(d-4)(d+4)=0(mod 27)
d^2=16(mod 27)
d=(+/-)4 (mod 27)
2d^2<21875
d<=104
So there're 16 values of d.
2N^2 is even, 350000 is even. So 3M^2 is even.
Let M=2a
12a^2+2N^2=350000
6a^2+N^2=175000
6a^2 is even, 175000 is even. So N^2 is even.
Let N=2b
6a^2+4b^2=175000
3a^2+2b^2=87500
2b^2 is even. 87500 is even. So 3a^2 is even.
Let a=2c
12c^2+2b^2=87500
6c^2+b^2=43750
6c^2 is even. 43750 is even. So b^2 is even.
Let b=2d
6c^2+4d^2=43750
3c^2+2d^2=21875
4c=M, 4d=N
If c =3k
3c^2=0(mod 27)
21875=5(mod 27)
So 2d^2=5(mod 27)
2d^2<21875
2d^2=27e+5
e is odd.
e<810
Let e=2f+1
2d^2=54f+32
d^2=27f+16
2f+1<810
f<=404
(d-4)(d+4)=27f
(d-4)(d+4)=0(mod 27)
d^2=16(mod 27)
d=(+/-)4 (mod 27)
2d^2<21875
d<=104
So there're 16 values of d.
由 ---- 於 星期二 五月 27, 2003 5:01 pm
or~~
Sub. M=100x, N=100y
After solving
3x^2+2y^2=35
then 倍大100 倍
Sub. M=100x, N=100y
After solving
3x^2+2y^2=35
then 倍大100 倍
由 ---- 於 星期二 五月 27, 2003 4:55 pm
I can never solve your second question, meowth, but for the first question,
3M^2+2N^2=35
If M=0(mod 3)
3M^2<35
M=3
3M^2=0(mod 27)
35=8(mod 27)
So 2N^2=8(mod 27)
N^2=4(mod 27)
2N^2<35
So N^2=4
N= (+/-)2
When M=3k(+/-)1,
3M^2=3(mod 9)
35=8(mod 9)
So 2N^2=5(mod 9)
Since 2N^2<35
2N^2=5, 14, 23, 32
N^2=16 (5/2, 7, 23/2 rejected)
N=(+/-)4
3M^2+2N^2=35
If M=0(mod 3)
3M^2<35
M=3
3M^2=0(mod 27)
35=8(mod 27)
So 2N^2=8(mod 27)
N^2=4(mod 27)
2N^2<35
So N^2=4
N= (+/-)2
When M=3k(+/-)1,
3M^2=3(mod 9)
35=8(mod 9)
So 2N^2=5(mod 9)
Since 2N^2<35
2N^2=5, 14, 23, 32
N^2=16 (5/2, 7, 23/2 rejected)
N=(+/-)4
由 --- 於 星期二 五月 27, 2003 4:14 pm
siuhochung 寫到:My method:
5x^2+16xy+14y^2=35
3(x+2y)^2+2(x+y)^2=35
My Q:
How many integer solutions for 3M^2+2N^2=35??
==>
Then, how many integer solutions for 3M^2+2N^2=350000 ??
由 --- 於 星期二 五月 27, 2003 4:09 pm
how
由 heron0520 於 星期二 五月 27, 2003 2:17 pm
see see
由 yptsoi 於 星期一 五月 26, 2003 8:22 pm
cc
由 Raceleader 於 星期一 五月 26, 2003 6:59 pm
yes 
由 Searchtruth 於 星期一 五月 26, 2003 6:58 pm
抱歉拉...剛剛在寫ㄉ答案...少ㄌ(+/-)..^^"
is stone me?
3Q~~^^
is stone me?
3Q~~^^
由 E.T 於 星期一 五月 26, 2003 6:45 pm
CC