Given that $(1 + kx)^4(1 + x)^n = 1 + 13x + 74x^2 + \cdots + k^4x^{n + 4}$, find the possible positive integer values of $k$ and $n$.
My working:
$$(1 + kx)^4(1 + x)^n = (1 + 4kx + 6k^2x^2 + 4k^3x^3 + k^4x^4)(1 + x)^n$$
This can be written as
$$(1 + x)^n + 4kx(1 + x)^n + 6k^2x^2(1 + x)^n + 4k^3x^3(1 + x)^n + k^4x^4(1 + x)^n$$
Therefore, the coefficient of $x$ in this expansion is
$$n + 4k = 13$$
Another equation in $n$ and $k$ is needed in order to solve. Hence we find the coefficient of $x^2$ in this expansion
$$\binom{n}{2} + 4kn + 6k^2 = 74$$
Furthermore, this simplifies to
$$n(n - 1) + 4k(2n + 3k) = 148 = 2^2 \times 37$$
Subbing $n = 13 - 4k$ we get
$$(13 - 4k)(12 - 4k) + 4k(11k + 26) = 2^2 \times 37$$
Which simplifies to
$$15k^2 + k + 2 = 0$$
This quadratic has non-integer roots so does not comply with the restrictions set by the question.
From here I am unable to progress. What is the next step in the solution?