or~~
Sub. M=100x, N=100y
After solving
3x^2+2y^2=35
then 倍大100 倍
[數學]求整數解?
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.
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---- - 訪客
