※ 引述 《hahaha2009》 之銘言：
: 請問各位大大..
:
: 已知a,b是正整數
: a^2+b^2除以a+b，商q餘r，並且q^2+r=1831，求(a,b)有幾組?
:
: 謝謝
r = 0 no solution, 0 < r < a+b
a = b implies r = 0
WLOG write a > b
Clearly (a+b)(a/2) < aa + bb < (a+b)a
Thus a/2 <= q < a
r < a+b < 2a <= 4q
qq < 1831 < qq + 4q < (q+2)^2
q < 42.xxx < q+2
q = 41, 42
r = 150, 67
Consider bb + (-q) b + (aa-aq-r) = 0
D = qq - 4(aa-aq-r)
= - (2a-q)^2 + 2qq + 4r = kk
b = q/2 +- k/2
2qq + 4r = (2a-q)^2 + kk
(Case 1) q = 41, r = 150
2qq + 4r = 2 * 7 * 283
Not a sum of square
(Case 2) q = 42, r = 67
2qq + 4r = 2^2 * 13 * 73
Is a sum of square
write k = 2m
13 * 73 = (a-21)^2 + mm
13 <-> 3 + 2i
73 <-> 8 +- 3i
13 * 73 <-> 18 + 25i, 30 + 7i
(a-21, m)
= +- (7,30), (18,25), (25,18), (30,7)
By a/2 <= q < a, we have 42 < q <= 84
(a-21, m)
= (25, +-18), (30, +-7)
Since b = 21 +- m
(a, b)
= (46, 21+-18), (51, 21+-7)
Since r = 67 < a+b
(a, b)
= (46, 39), (51, 28)
These are the 4 sol. of (a, b).
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