Let `P(n) = 7^(4^(th)) – 1` be divisible by `2^(2n+3)`
`P(1) = 7^(4) – 1 = (7^(2) – 1) (7^(2) + 1)`
` = 48 xx 50 = 32 xx 75`
`= 2^(5) xx 75`.
which is divisible by `2^(2xx1+3)`
Let us assume that the result is true for n` = k`.
i.e, `7^(4^(k)) – I` is divisible by `2^(2k+3)` but not by `2^(2k+4)`.
`rArr 7^(4^(k)) – 1 = 2^(2k+3)`m , where m is some odd natural number
Now, `7^(4^(k+1)) – 1 = (7^(4^(k)))^(4-1)`
`= ((2^(2k+3)m+1)^(2) +1)((2^(2k+3)m+1)^(2)-1)`
`= ((2^(2k+3)m+1)^(2) +1)(2^(2k+3)+2)(2^(2k+3)m)`
`= (2^(4k+6)m^(2) + 2^(2k+4) m + 2)(2^(2k+3)m + 2) (2^(2k+3)m)` ltbr `= 2^(2k-5)(2^(4k+5)m^(2) + 2^(2k+3) m + 1)(2^(2^(2k+2)m +1) (m),`
Which is divisivble by `2^(2k+5)`
Thus, `P(k + 1)` is true whenever `P(k)` is true.
So, by the principle of mathematical induction, `P(n)` is true for any natural number n.