Let n be an even positive integer such that $\frac{n}{2}$ is odd and let $α_{0}, α_{1}, ......... α_{n - 1}$ be the
complex roots of unity of order n .When $\prod_{k = 0}^{n - 1} (a + b α_{k}^{2}) = {(a^{\frac{n}{t}} + b^{\frac{n}{p}})}^{q}$ for any complex numbers a and b then $t + p + q =$ …..

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Detailed Solution

Let the complex number $α$ such that $α^{2} = \frac{a}{b}$
$f (x) = x^{n} - 1 = (x - α_{0}) (x - α_{1}) (x - α_{2}) ........ (x - α_{n - 1})$
We have $f (\frac{α}{i}) = {(\frac{1}{i})}^{n} . (α - i α_{0}) (α - i α_{1}) ....... (α - i α_{n - 1})$
$f (\frac{α}{i}) = f {(- \frac{1}{i})}^{n} . (α + i α_{0}) (α + i α_{1}) ....... (α + i α_{n - 1})$
Hence ,
$f (\frac{α}{i}) f (- \frac{α}{i}) = (α^{2} + a_{0}^{2}) (α^{2} - α_{1}^{2}) ........ (α^{2} - a_{n - 1}^{2}) = b^{n} \prod_{k = 0}^{n - 1} (α + α_{k}^{2}) b^{n} f (\frac{α}{i}) . f (- \frac{α}{i}) = b^{n} . {[{(a^{2})}^{2 m + 1} + 1]}^{2} = b^{n} {[{(\frac{a}{b})}^{2 m + 1} + 1]}^{2} b^{2 (2 m + 1)} {(\frac{a^{2 m + 1} + b^{2 m + 1}}{b^{2 m + 1}})}^{2} = (a^{n / 2} + b^{n / 2})$