I came across the following equation:
$$x=(10^ab+c)^{4d+1}-c$$
Why is $x$ a multiple of $10$ for any natural number values for $a$, $b$, $c$ and $d$?
The only progress I made was that $a$ could be removed from the equation, since $b$ can anyway be a multiple of $10$ or its powers.
On
Since natural number implies $a,b,c,d\ge1$ $$(10^ab+c)^{4d+1}=\sum_{k=0}^{4d+1}(10^ab)^kc^{4d+1-k}=c^{4d+1}+\underbrace{\sum_{k=1}^{4d+1}(10^ab)^kc^{4d+1-k}}_{\text{divisible by }10}$$ And since cyclicity of all natural numbers is a factor of 4,i.e. of $\{1,2,3,4,5,6,7,8,9\}$ are $\{1,4,4,2,1,1,4,4,2\}$; $c^{4d+1}$ ends in c and after subtracting c the number which now has last digit zero became divisible now by 10.
You can forget about $a$ and $b$ and concentrate on proving that $c^{4d+1}-c$ is a multiple of $10$.
Since $10=2\cdot 5$, you only need to prove that $c^{4d+1}-c$ is a multiple of both $2$ and $5$.
It is easy to see that $c^{4d+1}-c$ is a multiple of $2$ since even $\cdot$ even is even and odd $\cdot$ odd is odd.
It is easy to see that $c^{4d+1}-c$ is a multiple of $5$ when $c$ is a multiple of $5$ essentially because all multiples of $5$ end with either $5$ or $0$.
The crux of the matter is proving that $c^4-1$ is always a multiple of $5$ when $c$ is not a multiple of $5$. Write $c=5q+r$ with $1\le r \le 4$ and you'll only need to consider $r^4-1$, which is easily done in these four cases.