求a^2+b^2+c^2=a^2b^2的所有整数解

shingrons

没思路...
。。。。。

hamilan911

原帖由 shingrons 于 5-3-2007 11:33 PM 发表
没思路...
。。。。。

c^2 + 1 = a^2b^2 - a^2 - b^2 + 1
        = (a^2 - 1)(b^2 - 1)
since a^2 = 0 or 1(mod4)
a^2 - 1 = 0 or -1 (mod4) ,similar for b^2 - 1.
so RHS = 0 or 1 (mod4)
LHS = 1 or 2 (mod4)
for both side to be equal,a,b,c must be even number.
so let a = 2a' ,b = 2b' ,c = 2c'
so RHS = (2a'+1)(2a'-1)(2b'+1)(2b'-1)
so if a',b' are not the multiple of 3,then either 2a'+1 or 2a'-1 is divisible by 3,similar for 2b'+1,2b'-1.
so RHS = 0(mod9)
but c^2 = 0,1,4,7(mod9)
so LHS = 1,2,5,8(mod9)   -> no positive or negative integer solution
if a',b' are multiple of 3, then RHS = 1(-1)(1)(-1) = 1(mod3)
but 4c'^2 + 1 = 2(mod3)  -> no positive or negative integer solution
hence only solution is (a,b,c) = (0,0,0)
解得蛮长下
有其他解法吗？

dunwan2tellu

基本上，证明到 a,b,c 是偶数后，用无穷梯降法，得到 a=b=c=0

就好象你设 a=2a',b=2b',c=2c' 得到

a'^2 + b'^2 + c'^2 = 4a'^2 b'^2

从而 a',b',c' 又是偶数。然后又设她是 2a'' 的 pattern 又不断得到是偶数所以只有 当他们 = 0 时才有可能。

shingrons

hamilan911
你的解法我不是很明白
c^2 + 1 = a^2b^2 - a^2 - b^2 + 1
        = (a^2 - 1)(b^2 - 1)
since a^2 = 0 or 1(mod4)
a^2 - 1 = 0 or -1 (mod4) ,similar for b^2 - 1.
so RHS = 0 or 1 (mod4)
这里都ok，下面
LHS = 1 or 2 (mod4)
for both side to be equal,a,b,c must be even number.
我不懂
可以解释一下下？

hamilan911

原帖由 shingrons 于 8-3-2007 11:24 PM 发表
hamilan911
你的解法我不是很明白
c^2 + 1 = a^2b^2 - a^2 - b^2 + 1
        = (a^2 - 1)(b^2 - 1)
since a^2 = 0 or 1(mod4)
a^2 - 1 = 0 or -1 (mod4) ,similar for b^2 - 1.
so RHS = 0 or 1 (mod4)
...

RHS = 0 or 1(mod 4)
LHS = 1 or 2(mod4)
for both side to equal,u get RHS = LHS = 1(mod4)  -(1)
等同于证明 (1)
LHS = 1(mod4) 等同于证明 c^2 = 0(mod4)   hence c is even number.
RHS = 1(mod4) 等同于证明 a^2 - 1 = -1(mod4)  ,  b^2 - 1 = -1(mod4)
也等同于证明 a^2 = 0(mod4) , b^2 = 0(mod4) 利用同理可得 a,b is even number.