## 帳號 自動登入 取回密碼 密碼 註冊

# 請教數對解題 發表於 08-10-25 18:00:47 |顯示全部樓層 大字 中字 小字  發表於 08-10-25 18:26:24 |顯示全部樓層 大字 中字 小字 載入全部圖片
 a^2+b^2=(51-10)(51+10)=51^2-10^2=(50+1)^2-10^2=50^2+2*50+1-10^2=50^2+1^2 (a,b)=(50,1)or(1,50)

http://ej0cl6.pixnet.net/blog 發表於 08-10-26 12:33:42 |顯示全部樓層 大字 中字 小字 載入全部圖片
 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) 發表於 08-10-26 12:34:07 |顯示全部樓層 大字 中字 小字 載入全部圖片
 底下是一位老師寫下的解答 -------------- 令a+b=k ab=(k^2-2501)/2 a,b為 X^2-kX+(k^2-2501)/2=0兩根 D=5002-k^2必須為完全平方數 so 2501

 你需要登入後才可以回覆 登入 | 註冊 回覆並轉播 回覆後切換到最後一頁

GMT+8, 20-12-2 01:50 , Processed in 2.295028 second(s), 16 queries .