A maths MI

let T(n) = 3^(2n+1) + 2^(n+2)
T(1) = 35 = 5 * 7
T(2) = 259 = 7 * 37
so, p should be 7

T1) is true
suppose T(k) is true, i.e. 3^(2k+1) + 2^(k+2) = 7m where m is integer
T(k+1) = 3^( 2(k+1)+1 ) + 2^( (k+1)+2 )
= 3^( 2k+3 ) + 2^( k+3 )
= 9 * 3^( 2k+1 ) + 2 * 2^( k+2 )
= 9 * ( 7m - 2^(k+2) ) + 2 * 2^( k+2 )
= 9 * 7m - 7 * 2^( k+2 )
= 7 * ( 9m - 2^( k+2 ) )
is multiple of 7

so, proof completed

(a) The prime value is 7.
(b) Let P(n) be the statement that:
32n+1 + 2n+2 is divisible by 7
When n = 1,
32n+1 + 2n+2 = 35
Then P(1) is true.
Suppose that P(k) is true where k is a positive integer, i.e.

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May07/MI5.jpg

Therefore P(k+1) is also true.
By the principle of MI, P(n) is true for all positive integers n.

因為 3^(2n+1)+2^(n+2) = 3^3 + 2^3 = 35 when n = 1
因為 3^(2n+1)+2^(n+2) = 3^5 + 2^4 = 259 when n = 2

所以p 只可能是gcd(35,259) = 7
設S(n)是斷句存在整數m 3^(2n+1)+2^(n+2) = 7m

S(1)是正確的, m = 5

假設S(k)是正確的
3^(2k+1)+2^(k+2) = 7m

3^(2(k+1)+1)+2^((k+1)+2)

9(3^(2k+1))+2(2^(k+2))
= 9(3^(2k+1))+2(2^(k+2)) + 7(2^(k+2)) - 7(2^(k+2))
= 9(3^(2k+1))+9(2^(k+2)) - 7(2^(k+2))
= 9(3^(2k+1))+(2^(k+2))) - 7(2^(k+2))
= 7m - 7(2^(k+2))
= 7n

所以S(k+1)也正確。
所以對於所有自然數n而言，S(n)是正確的。