An algebra problem by mietantei conan

If we write $N_\pm = \big(3 \pm \sqrt{7}\big)^{20}$ (so that $N = N_+$ ), then $N_\pm \; = \; \sum_{j=0}^{20} \binom{20}{j} 3^{20-j}\big(\pm\sqrt{7}\big)^j$ and hence $N_+ + N_- \; =\; 2\sum_{j=0}^{10} \binom{20}{2j} 3^{20-2j} 7^j$ is a positive integer. Now $0 < 3 - \sqrt{7} < 1$ , and hence $0 < N_- < 1$ , so it follows that $f \,=\, 1 - N_-$ , and hence $N(1-f) \; = \; N_+ N_- \; = \; (3^2 - 7)^{20} \; = \; 2^{20} \; = \; 1048576$ making the answer $\boxed{576}$ .