高一數學題-急~(有關於質數的問題),萬分感謝!

Question

若n為質數,且k=(n^2-3n+38)/(n-1)亦為質數,求數對(n,k)=?

Answer 1

(n^2 - 3n + 38) / (n-1) =
n - 2 + (36/(n-1))

To be an integer,
n-1 = 1, 2, 3, 4, 6, 9, 12, 18, 36
so n = 2, 3, 4, 5, 7, 10, 13, 19, 37
But n is prime,
so n = 2, 3, 5, 7, 13, 19, 37
and k is 36, 19, 12, 11, 14, 19, 36 respectively.

But k is prime, so k = 19, 11, 19
And the answer is (n,k)=(3, 19), (7, 11), or (19,19)

倉促中也許有誤.

2007-09-13 08:24:11 補充：
因為發現都沒人回答,
我為了搶第一個回答, 倉促中寫出答案.
現在經過檢查之後,
確定應該無誤.

Answer 2

(n^2-3n+38)/(n-1)
=(n+2)+〔36／(n-1)〕
因(n+2)是整數，故〔36／(n-1)〕是整數

n-1｜36
故n-1=1,2,3,4,6,9,12,18,36
n=2,3,4,5,7,10,13,19,37（4,10不行）
代回　k
n=2時k=36(不合)
n=3時k=19
n=5時k=12(不合)
n=7時k=11
n=13時k=14(不合)
n=19時k=19
n=37時k=36(不合)
故共有3對(3,19)(7,11)(19,19)