In  $n$ is the positive integer, then the integral part of ${(7+5\sqrt{2})}^{2n+1}$, is

$I+f={(5\sqrt{2}+7)}^{2n+1},g={(5\sqrt{2}-7)}^{2n+1}$

$I+f-g=2{[}^{(2n+1)}{C}_1{\left(5\sqrt{2}\right)}^{2n}{\left(7\right)}^1+{}^{(2n+1)}{C}_3{\left(5\sqrt{2}\right)}^{2n-2}\cdot {\left(7\right)}^3+\cdots ]$

$I+f-g=$ even integer $\to \left(i\right)$

Here, $0<f<1,0<g<1\Rightarrow -1<f-g<1$

As  $(f-g)$ is an integer between $-1$ to 1

$\therefore f-g=0\to \left(ii\right)$

From equations $\left(i\right)\text{\hspace{0.17em}and\hspace{0.17em}}\left(ii\right):I=$ even integer.