if $(6 \sqrt{6} + 14)^{2 n + 1} = p$ , prove that the integral part of P is an even integer and PF = 20 $^{2 n + 1}$ where F is the fractional part of P

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Solution

Let $(6 \sqrt{6} + 14)^{2 n + 1}$ = P = I + F
where I is a positive integer and 0 < F < 1
clearly $(6 \sqrt{6} - 14)^{2 n + 1}$ = F' where 0 < F' < 1
Subtracting, we get
$2 [^{2 n + 1} C_{1} (6 \sqrt{6})^{2 n} \times 14 +^{2 n + 1} C_{3} (6 \sqrt{6})^{2 n - 2} \times 14^{3} + . . . . .]$
= I + F - F'
$∴$ I + F - F' = an even integer.
But since 0 < F < 1 and 0 < F' < 1, the only possible is that F - F' = 0 or F = F'.
Hence I is an even integer.
Also PF = PF' = $(6 \sqrt{6} + 14)^{2 n + 1} (6 \sqrt{6} - 14)^{2 n + 1} = (216 - 196)^{2 n + 1} = 20^{2 n + 1} .$

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if (6√6+14)2n+1=p , prove that the integral part of P is an even integer and PF = 202n+1 where F is the fractional part of P

if $(6 \sqrt{6} + 14)^{2 n + 1} = p$ , prove that the integral part of P is an even integer and PF = 20 $^{2 n + 1}$ where F is the fractional part of P