兩到高一數學證明題

1 Consider (n^2-13)/(n+5) if it is in simplest form. Then
(n+5,n^2-13)=1=>(n+5,n^2-13-n(n+5)=1
=>(n+5,5n+13)=1=>(n+5,5n+13-5(n+5))=1=>(n+5,12)=1
Since 12=2*2*3, we conclude that n+5 is coprime with 2 and 3
From 0<n<2003, the number of n+5 which is coprime with 2 and 3 is
2007-[2007/2]-[2007/3]+[2007/6]-2 (since we omit 1,5)
=2007-1003-669+334-2
=667
There are 667 different values of n
The general formula is
(N+5)-[(N+5)/2]-[(N+5)/3]+[(N+5)/6]-2
2 if this is not true, then AB+1=MA=>AB-MA=-1
=>A(M-B)=1
That means the product of two integers is equal to 1, which is a contradiction !

2009-08-16 12:34:04 補充：
For Q.2, we assume A>1 !

1.(n^-13,n+5)=1
(n^-13)/(n+5)=n+(-5n-13)/(n+5)
=n-5+[12/(n+5)]
所以若此為最簡分數，則n+5不為2，3之倍數
故所求=1335個

2.設(ab+1，a)=d
則d l ab+1且d l a
d l ab+1-ab
d l 1
d=1
故ab+1與a互質

算錯請指正