+505 Using euclidean's algorithm, we see that
\(\gcd(n^2, n-12)\\=\gcd(n^2-n(n-12), n-12)\\=\gcd(12n, n-12)\\=\gcd(12n-12(n-12), n-12)\\=\gcd(144,n-12)\)
Therefore, 144 must be divisible by n-12 in order for n^2 to be divisible by n-12.
\(144=2^43^2\), so it has (4+1)(2+1) = 15 factors (which are \(2^03^0, 2^13^0, 2^23^0, 2^33^0, 2^43^0, 2^03^1, 2^13^1, 2^23^1, 2^33^1, 2^43^1, 2^03^2, 2^13^2, 2^23^2, 2^33^2, 2^43^2\)), which means that if you add 12 to any of the numbers mentioned above, you will get a number n that satisfies the condition above.
Can you finish it?
