Form4 Maths

Let P(n) be 5^(2n-1)-3^(2n-1)-2^(2n-1) is divisible by 15 for all positive integers n.

When n = 1,
5^(2*1-1)-3^(2*1-1)-2^(2*1-1)
=5-3-2
=0
which is divisible by 15
So P(1) is true

Assume P(k) is true, i.e. 5^(2k-1)-3^(2k-1)-2^(2k-1) is divisible by 15, i.e.
5^(2k-1)-3^(2k-1)-2^(2k-1) = 15m for a positive integer m

When n = 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^(2k-1+2)-3^(2k-1+2)-2^(2k-1+2)
=(5^2)(5^(2k-1)) - (3^2)(3^(2k-1)) - (2^2)(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))
=15(4m) + 7*3*5*(5^(2k-2)) - 5*3*(3^(2k-1))
=15(4m) + 15(7*5^(2k-2)) - 15(3^(2k-1))
=15(4m + 7*5^(2k-2) - 3^(2k-1))
which means it's divisible by 15
So P(k+1) is true.

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

When n = 1,
5^(2*1-1)-3^(2*1-1)-2^(2*1-1)
=5-3-2
=0
which is divisible by 15
So P(1) is true

Assume P(k) is true, i.e. 5^(2k-1)-3^(2k-1)-2^(2k-1) is divisible by 15, i.e.
5^(2k-1)-3^(2k-1)-2^(2k-1) = 15m for a positive integer m

When n = 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^(2k-1+2)-3^(2k-1+2)-2^(2k-1+2)
=(5^2)(5^(2k-1)) - (3^2)(3^(2k-1)) - (2^2)(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))
=15(4m) + 7*3*5*(5^(2k-2)) - 5*3*(3^(2k-1))
=15(4m) + 15(7*5^(2k-2)) - 15(3^(2k-1))
=15(4m + 7*5^(2k-2) - 3^(2k-1))
which means it's divisible by 15
So P(k+1) is true.

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