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一個頗長的方法...
a^2 + b^2 = 41*61
41*61 is an odd number ==> one of a and b must be and odd number and the other must be even.
observe that if (x,y) is a solution, (y,x) is also a solution.
Without loss of generality, we can assume a is odd and b is even
then we can write a=2n+1 and b=2m, for some non negative integer n and positive integer m.
(2n+1)^2 + (2m)^2 = 2501
4n^2 + 4n + 4m^2 = 2500
n^2 + n + m^2 = 625
n^2 + n is always even and 625 is odd ==> m is odd
let m = 2k+1 for some non negative integer k
n^2 + n + (2k+1)^2 = 625
n^2 + n + 4k^2 + 4k = 624
4k^2 + 4k and 624 are divisible by 8
==> n^2 + n is divisible by 8
==> n = 0 or -1 ( mod 8 )
if n = 0 ( mod 8 )
let n = 8h for some non negative integer h
(8h)^2 + 8h + 4k^2 + 4k = 624
16h^2 + 2h + k^2 + k = 156
Since k is non negative, the possible values of h are 0, 1, 2, 3
( otherwise, 16h^2 + 2h + k^2 + k > 156)
substitute the values of h into the equation and solve for k
the only integer solutions of (h,k) are: (0,12) or (3,2)
==> (a,b) = (1,50) or (49,10)
from the observation above, (a,b) can also be (50,1) or (10,49)
if n = -1 ( mod 8 )
let n = 8h-1
(8h-1)^2 + 8h-1 + 4k^2 + 4k = 624
16h^2 - 2h + k^2 + k = 156
Again, since k is non negative, the possible values of h are 0, 1, 2, 3
( otherwise, 16h^2 - 2h + k^2 + k > 156)
substitute the values of h into the equation and solve for k
the only integer solution of (h,k) is: (0,12) (again......)
so, the only solutions of (a,b) are (1,50) or (49,10) or (50,1) or (10,49) |
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發表於 08-10-25 18:00:47
