76. Given that for any positive integer $\mathrm{n}, 9^{2 n}-8^{2 n}-17$ is always divisible by $\mathrm{m}$, find the largest positive integer $m$.


Solution: Since $9^{2 n}-8^{2 n}-17=9^{2 n}-\left(8^{2 n}-1\right)-18=9^{2 n}-\left(64^{n}-1\right)-18$,
because $9^{2 n}, 18$ are both divisible by 9,
and $64^{n}=(63+1)^{n}=(63+1) \times(63+1) \times \cdots \times(63+1)$, in the expansion, except for 1, all other terms are multiples of 63, so $(63+1)^{n}-1$ is also divisible by 9.
Therefore, $9^{2 n}-8^{2 n}-17$ is divisible by 9.
Since
$$
9^{2 n}-8^{2 n}-17=\left(81^{n}-1\right)-8^{2 n}-16 \text {, }
$$

similarly, $9^{2 n}-8^{2 n}-17$ is divisible by 16.
Since
$$
9^{2 n}-8^{2 n}-17=\left(81^{n}-64^{n}\right)-17 \text {, }
$$

so $9^{2 n}-8^{2 n}-17$ is divisible by 17.
Since the least common multiple of $9,16,17$ is 2448,
$9^{2 n}-8^{2 n}-17$ is divisible by 2448.
When $n=1$, $9^{2 n}-8^{2 n}-17=0$, and when $n=2$, $9^{2 n}-8^{2 n}-17=2448$. Therefore, $\mathrm{m}$ is at most 2448.