Question

Prove that for any natural n:

a

the number 7^(n+4) −7^n is divisible by 30.

b

the number 3^(n + 2) −2^(n +2) +3^n −2^n is divisible by 10.

Answer 1

a) Let
`n=0`; then
`7^4-1=2400=30\cdot80`, which is divisible by 30. We have

`7^(n+4)-7^n=7^n(7^4-1)`

and since we know
`30\mid7^4-1`, it follows that
`30\mid7^(n+4)-7^n` for all
`n\ge1`.

b) We can write

`3^(n+2)-2^(n+2)+3^n-2^n=(3^(n+2)+3^n)-(2^(n+2)+2^n)`

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

`=10\cdot3^n+5\cdot2^n`

`=10\cdot3^n+10\cdot2^(n-1)`

`=10\left(3^n+2^(n-1)\right)`

and we're done.