<s>[INST] Powers and differences of positive integers Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5=b^4$, $c^3=d^2$, and $c-a=19$. Determine $d-b$.
I know this question isn't particularly hard, but I've been having trouble getting any substantial ground. I've tried substituting that $c=a+19$ and $a=c-19$ into the first two equations, but it hasn't looked very useful.
A path to a solution or a few hints would be great. Thanks!
 [/INST] From $a^5 = b^4 \implies a = \left(\frac{b}{a}\right)^4$, we can deduce that $\frac{b}{a}$ is an integer. I will let $m = \frac{b}{a}$.
Similarly, from $c^3 = d^2 \implies c = \left(\frac{d}{c}\right)^2$, we can deduce that $\frac{d}{c}$ is an integer. I will let $n = \frac{d}{c}$. 
Since $a = m^4$ and $c = n^2$, we have
$$n^2 - m^4 = 19$$
$$(n + m^2)(n - m^2) = 19$$
Since $19$ is a prime, we only have to check the cases for which $19 = 1\cdot19$ and for which $19 = -1\cdot-19$. If we let $n + m^2 = 19$ and $n - m^2 = 1$, then we can get $n = 10, m = 3$. I suppose it is trivial to show that all other cases do not give valid results.
Hence, $\frac{d}{c} = n = 10 \implies d = 10c$, and $\frac{b}{a} = m = 3 \implies b = 3a$. It should be easy to continue on from here!
(Just in case you wanna check: $a = 81, b = 243, c = 100, d = 1000$)
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