Gerry Myerson shows that this is a sum-of-two-squares problem.
This is related to: Efficiently finding two squares which sum to a prime
Given a^2 * b^2 * c, we can efficiently find m, n such that m^2 + n^2 = a^2 * b^2 * c. Then it's a simple matter of checking to see if a or b divides m, and if so, if the other value divides n.
I believe an outline of the proof process you want to go through, goes something like this:
Let X = a^2 * b^2 * c
If X is divisible by a prime equal to 3 mod 4, then X is divisible by that prime raised to an even power, and both 'm' and 'n' will be divisible by that prime raised to that even power. So we can effectively factor out those primes.
We now have Y divides a^2 * b^2 * c and Y is a product of primes equal to 1 mod 3. And we can look for 'k' and 'l' such that k^2 + l^2 == Y.
If Y is prime, then there is a unique solution, and the above referenced stackexchange page discusses how to find 'k' and 'l'.
If Y is not prime, then we can find a solution for each factor, and combine the partial solutions to enumerate all solutions. Suppose for example that Y = WZ where W and Z are prime and i^2 + j^2 = W and q^2 + r^2 = Z.
Then (iq + jr)^2 + (ir - jq)^2 = WZ. There are also solutions involving the conjugates i^2 + (-j)^2 and q^2 + (-r)^2. Equivalent solutions can be avoided by using only the positive form for one of the primes, but allowing either the positive form or the conjugate form for each of the other primes.
This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\
&=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right)
\end{align}$$
Where the last equation is because when $p$ is odd, $$
\begin{align}u^p+1&=(u+1)
\sum_{j=0}^{p-1} (-1)^ju^j
\\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$
Now, since $n>d,$ we can set $$
\begin{align}x&=a^{n-d}t^{k-p}\\
y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}
\end{align}$$
For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.
So this solves all cases $k>3.$
You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$
k even
So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$
Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$
k odd
Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$ Then for any integer $t$:
$$\begin{align}z&=a^{m-1}t^{2m-1}\\
x&=at^2\\
y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1}
\end{align}$$
is a solution.
In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$
Best Answer
As @Qiaochu Yuan noted, $10^{n-\log p}=\dfrac{10^n}{p}$ is always a rational number.
Suppose that a positive integer $n$ and an integr $p$ are given, and let $a$ and $b$ to be integers satisfying the above relation:
$$ \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{10^n} \Longleftrightarrow \\ pab = 10^na + 10^nb \Longleftrightarrow \\ p^2ab = 10^npa + 10^npb \Longleftrightarrow \\ p^2ab - 10^npa - 10^npb = 0 \Longleftrightarrow \\ p^2ab - 10^npa - 10^npb + 10^{2n} = 10^{2n} \Longleftrightarrow \\ (pa - 10^n) (pb - 10^n) = 10^{2n} . $$
notice that both of the factors $(pa - 10^n)$ and $(pb - 10^n)$ are (positive or negative) divisor of $10^{2n}$, so we can conclude that there is a (positive or negative) divisor $d$ of $10^{2n}$; such that:
$$ (pa - 10^n)=d \ \ \ \ \ \ \text{and} \ \ \ \ \ \ (pb - 10^n)=\dfrac{10^{2n}}{d} \ \ \ \ \ \ \Longleftrightarrow $$
$$ a = \dfrac {d+10^n} {p} \ \ \ \ \ \ \text{and} \ \ \ \ \ \ b = \dfrac { \dfrac{10^{2n}}{d} +10^n} {p} \ \ \ \ \ \ . $$