Question

Prove by induction that `f(n)=2^(2n-1)+3^(2n-1)` is divisible by 5 for `n in ZZ^+`?

Answer 1

See below.

Explanation:

Note that for `m` odd we have

`(a^m+b^m)/(a+b) = a^(m-1)-a^(m-2)b+ a^(m-3)b^2 + cdots -a b^(m-2)+b^(m-1)`

which demonstrates the afirmation.

Now by finite induction.

For `n = 1`

`2+3 = 5` which is divisible.

now supposing that

`2^(2n-1)+3^(2n-1)` is divisible we have

`2^(2(n+1)-1)+3^(2(n+1)-1) =2^(2n-1) 2^2+3^(2n-1)3^2=`
`= 2^(2n-1) 2^2+3^(2n-1)2^2+5 xx 3^(2n-1)=`

`= 2^2(2^(2n-1)+3^(2n-1))+5 xx 3^(2n-1)` which is divisible by `5`

so it is true.