Re: 國安局數論考古題
※ 引述《bienhayes ()》之銘言:
: Let u,v 自然數, 滿足 (a) u^2 + 16^2 = v^2 和 (b) i^u + i^v = -2,
: 其中 i 是根號 -1. 求所有 (u,v) 之值.
: 大概寫一下我的答案:
: 由 (b) 知 u and v are congruent to 2 (mod 4),
: write u = 4a+2, v = 4b+2, where a and b are nonnegative integers.
: 代入 (a), 2^8 = (u+v)(v-u) = (4a+4b+4)(4b-4a) = 2^4 (a+b+1)(b-a)
: 得 16 = (a+b+1)(b-a).
到此OK
Since a and b are nonnegative integers, then a+b+1>b-a
and 16 = (a+b+1)(b-a)
Case I: a+b+1=8 and b-a=2 => a=2.5, b=4.5 not integer
Case II: a+b+1=16 and b-a=1 => a=7, b=8 => (u,v)=(30,34)
: if 2|(b-a), then b = 2c+a for some integer c.
: then a+b+1=2c+2a+1>0 is odd, which implies a=b=0, a contradiction.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
不懂你這行的理由是什麼
: hence 16 = a+b+1 and b-a = 1, which implies a=7 and b=8.
: so (u,v) 只有一解 (30,34).
: 這題被扣16分簡直是被KO,麻煩大大幫我看看 感謝!
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