F4 MI (20 pts ) diffcult

let S(n) be the proposition,
"5^(2n-1) - 3^(2n-1) - 2^(2n-1) is divisible by 15 for all positive integers n"
let f(n)=5^(2n-1) - 3^(2n-1) -
.
when n=1, f(1)=5^(2‧1-1) - 3^(2‧1-1) - 2^(2‧1-1)
....................=5-3-2
....................=0(15)
since f(1) is divisible by 15
therefore, S(1) is true
.
Assum S(k) is true, where k is a natural number, i.e.
5^(2k-1) - 3^(2k-1) - 2^(2k-1)=15m, where m is an natural number
.
when n=k+1, f(k+1)=5^[2(k+1)-1] - 3^[2(k+1)-1] - 2^[2(k+1)-1]
...........................=5^(2k+1)-3^(2k+1)-2^(2k+1)
...........................=(5^2) 5^(2k-1) - (3^2) 3^(2k-1) - (2^2) 2^(2k-1)
...........................=4[5^(2k-1) - 3^(2k-1) - 2^(2k-1)] + (21) 5^(2k-1) - (5) 3^(2k-1)
...........................=4(15m)+15[7‧5^(2k-2) - 3^(2k-2)]
...........................=15[4m+ 7‧5^(2k-2) - 3^(2k-2)]
since m, k are natural numbers
therefore [4m+ 7‧5^(2k-2) - 3^(2k-2)] must be a natural number
therefore, f(k+1) is divisible by 15 if S(k) is true
thus, S(k) is true, then S(k+1) is also true
By the principle of mathematical induction, S(n) is true for all natural numbers n

When n = 1, the expression is 5^1 - 3^1 - 2^1 = 0 which is divisible by 15 for all positive integers n.
So the result is true for n = 1.

Suppose the result is true for n = k,
that is 5^(2k-1) - 3^(2k-1) - 2^(2k-1) is divisible by 15.
So we can write 5^(2k-1) - 3^(2k-1) - 2^(2k-1) = 15M where M is an integer.

Consider 5^[2(k+1)-1] - 3^[2(k+1)-1] - 2^[2(k+1)-1]
= 5^(2k+1) - 3^(2k+1) - 2^(2k+1)
= 25 [5^(2k-1)] - 9 [3^(2k-1)] - 4 [2^(2k-1)]
= 4 [5^(2k-1) - 3^(2k-1) - 2^(2k-1)] + 21 [5^(2k-1)] - 5 [3^(2k-1)]
= 4 (15M) + 21 [5^(2k-1)] - 5 [3^(2k-1)]

Since k is a positive integer, 2k-1 is also a (odd) positive integer.
So 21 [5^(2k-1)] is divisible by both 3 and 5 and so divisible by 15 and we can write 21 [5^(2k-1)] = 15A, where A is an integer
Also 5 [3^(2k-1)] is divisible by both 3 and 5 and so divisible by 15 and we can write 5 [3^(2k-1)] = 15B, where B is an integer

Coming back to the question,
= 4 (15M) + 15A + 15B
= 15 (4M + A + B)
which is divisible by 15.
So, the result is also true for n = k+1.

By the principle of Mathematical Induction, 5^(2n-1) - 3^(2n-1) - 2^(2n-1) is divisible by 15 for all positive integers n.