How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$? How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$ ?
MY WORK :-
Well we know that $a+b$ divides $a^3+b^3$ thus $n+7$ divides $n^3+7^3$ thus it divides $n^3+343$. Now if $n+7$ divides $n^3+72$ then it divides their difference. Thus $n+7$ divides $343-72=271$. Now the factor of $271$ are $1$ and $271$ only. So if $n\lt100$ then there are no values possible; hence answer should be $0$. However in answer key it's said $1$. I want to know why it's $1$ ??


Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^2$.

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1, 
   10^2}] //. {} -> Nothing

Running the code gives:
Out[1]=Nothing

What implies that there are no solutions. If I extend the search to $1\le\text{n}\le10^8$ I found:
In[2]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1, 
   10^8}] //. {} -> Nothing

Out[2]={264}

Which means that $\text{n}=264$ is a solution, because:
$$\frac{264^3+72}{264+7}=67896\tag1$$
