Firstly, the value of  $(5\sqrt{2}+7)^{2n+1}-(5\sqrt{2}-7)^{2n+1}$ must be an even integer. This can be shown by expanding both terms and note that all the odd powers of $5\sqrt{2}$ are cancelled and the the terms doubled. Now because $$0 < (5\sqrt{2}-7)^{2n+1} < 1 \implies D=(5\sqrt{2}-7)^{2n+1}$$

Meanwhile, as $I+D=(5\sqrt{2}+7)^{2n+1}$ itself. So $$(I+D)D=(5\sqrt{2}+7)^{2n+1}(5\sqrt{2}-7)^{2n+1}=1$$

which is a constant.