Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.
Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$
Define $S(a,m)=1^m+2^m+3^m+…+a^m$ where $a,m\in\mathbb{Z}_+$
Edit:–Update claim
For $m$ is every positive integer. How do you show that
If $a\nmid S(a-1,m)$
Then $D(a,a^{m+1}-S(a,m))+D(a,S(a-1,m))=(a-1)(m+1)$?
And
If $a\mid S(a-1,m)$
Then $D(a,a^{m+1}-S(a,m))+D(a,S(a-1,m))\ne(a-1)(m+1)$?
Source code
n1= 2
o = 1
while n1 < 100:
m = 2
print("\n n1=",n1)
#print("m=",m)
num=n1
sum_num = 0
for i in range(1, num):
sum_num += i**(m)
n2 = (sum_num)
if(n2%num == 0):
print("div & sum=",n2)
else:
print("not div & sum=",n2)
rem_array = []
while n2 != 0:
mod = n2%n1
if mod != 0:
rem = mod
n2 = n2 - rem
rem_array.append(round(rem))
n2 = n2/n1
else:
n2 = n2/n1
rem_array.append(0)
print(rem_array[::-1],sum(rem_array))
# print(sum(rem_array))
n2=(sum_num)+n1**m
rem1_array = []
while n2 != 1:
mod = n2%n1
if mod != 0:
rem1 = n1-mod
n2 = n2 + rem1
rem1_array.append(round(rem1))
n2 = n2/n1
else:
n2 = n2/n1
rem1_array.append(0)
# print(rem_array)
print(rem1_array[::-1],sum(rem1_array))
if((n1-1)*(m+1) == sum(rem_array)+sum(rem1_array)):
print("oooooooooook")
print("(a-1)(m+1)=",(n1-1)*(m+1))
n1 += o
Update claim helps to show below claim
For $m$ is even positive integer. How do you show that
If $(a-1)\mid S(a-1,m)$
Then $D(a,a^{m+1}-S(a,m))+D(a,S(a-1,m))=(a-1)(m+1)$?
We can easily prove for every prime$-p>m+1$ is $p\mid S(p,m)$
Formula
$$ S(a,n)= \sum_{i=1}^{a} i^{n}=\sum_{b=1}^{n+1} \binom{a}b\sum_{j=0}^{b-1} (-1)^{j}(b-j)^{n}\binom{b-1}j$$
Proof
Let $a=p(prime)>n+1$
We can see, $a$ can be common out from $\sum_{b=1}^{n+1}\binom{a}b\sum_{j=0}^{b-1} …$
$\implies a|S(a,n)$
Best Answer
Your equation $D(a,a^{m+1} - S(a,m)) + D(a,S(a-1,m)) = (a-1)(m+1)$ is not always true when $(a-1) \mid S(a-1,m)$.
For example, take $a=4$, $m=3$. $S(3,3) = 1^3 + 2^3 + 3^3$ is divisible by $3$, but $D(4,4^4 - S(4,3)) + D(4,S(3,3)) = 6+3=9$ while $(4-1)(3+1) = 12$.