How many postive integer $n$ are there, such that $\frac{(n-1)^2}{n+29}$ is an integer? my solution $$\frac{n^2 - 2n +1}{n+29}\overbrace{\Rightarrow}^{\text{long division}} n -31 + \frac{900}{n+29}$$
Now the question is how many divisors for 900
$$900 \underbrace{=}_{\text{prime factorization}} 2^23^25^2$$
so the number of divisors is $(2+1)(2+1)(2+1) = 3^3 = 27$.
Is my solution correct?

Following along the lines of my comment as well as Stephen's, the factors of $900$ are:
$1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, 900$
However, in order to answer the question, the factors have to be rewritten in the form $n + 29$ such that $n$ is positive. This happens from $30$ and above.
Counting the number of factors of $900$ that are above $30$, we get $\fbox{13}$ different values of $n$ that are positive integers such that $\frac{(n - 1)^2}{n + 29}$ is also a positive integer.
