COMP SCI 241 Discrete Mathematics I
Homework # 22
Fall 2006
F. Baulieu
1) Use mathematical induction to prove: If n is any non-negative integer, then
53n  33n is a multiple of 7.
Pf (by induction on n)
3n
3n
0
0
Base case: If n = 0, then 5  3  5  3  1  1  0  7(0)
which is a multiple of 7. QED base case.
3n
3n
Inductive case: Suppose 5  3 is a multiple of 7 for some non-negative integer n.
3( n 1)
We must show that 5
 33(n 1) is a multiple of 7.
3n
3n
3n
3n
Since 5  3 is a multiple of 7, there is an integer t such that 5  3  7 t .
3n
3n
Thus 5  7 t  3 .
Now consider that
3( n 1)
3( n 1)
3n  3
3n  3
5
3
5
3
 5353n  3333n
 125  53n  27  33n
 125  (7 t  33n )  27  33n
 125  (7 t )  125  33n  27  33n
 125  (7 t )  98  33n
 7(125 t  33n )
3n
and since n is an integer, so is 3n, so 3 is an integer, and t is an integer, so 125t is an
3( n 1)
3n
 33(n 1) is
integer. So (125 t  3 ) is an integer, and we have shown that 5
a multiple of 7. QED
2) Use mathematical induction to prove: If n is any non-negative integer, then
52n 1  32n is a multiple of 4.
Pf (by induction on n)
2n 1
Base case: If n = 0, then 5
 32 n  5  1  4
which is a multiple of 4. QED base case.
Inductive case: Suppose 5
2n 1
 32n is a multiple of 4 for some non-negative integer
n.
We must show that 5
2(n 1) 1
 32(n 1) is a multiple of 4.
2n 1
Since 5
 32n is a multiple of 4, there is an integer t such that 52n 1  32n  4t
2n 1
Thus 5
 4t  32n
Now consider that
2( n 1) 1
2( n 1)
2n  3
2n  2
5
3
5
3
 535 2n  3232n
 125  5 2n  9  32n


 125 4 t  32n  9  32n
 1254 t   125  32n  9  32n
 4(125 t )  116  32n
 4(125 t  29  32n )
2n
and since n is an integer, so is 2n, so 3 is an integer, and t is an integer, so 125t is an
2n
integer. So (125 t  29  3 ) is an integer, and we have shown that
52(n 1) 1  32(n 1) is a multiple of 4.
QED