find all $ 1\le a\le b \le 100$，s.t. $1+ab\mid1+a^4$

Question

$1\le a \le b \le 100$ ，find all $a,b$ satisfy $1+ab\mid 1+a^4$ at first I think $b=a^3$ but I find $a=8$, $b=30$ .i think maybe I loss some $(a, b)$

Answer 1

A simple program gives the solutions below. It seems that you haven't missed anything. $$ \begin{array}{r|rrrrr} a & 1 & 2 & 3 & 4 & 8 \\ b & 1 & 8 & 27 & 64 & 30 \\ \end{array} $$ The solutions $1\le a\le b \le 5000$ with $b \ne a^3$ are $$ \begin{array}{r|rrrrr} a & 8 & 27 & 30 & 64 & 112 & 125 & 240 & 418\\ b & 30 & 240 & 112 & 1020 & 418 & 3120 & 2133 & 1560 \end{array} $$ These values of $a$ form a subsequence of OEIS/A115169. If we remove the restriction $a\le b$, then it seems we get the whole sequence.

Answer 2

Here is a Python code to check such pairs of $a$ and $b$ which satisfies the divisibility condition and $b\neq a^3$.

for a in range(1,101):       # range(n,m) works upto m-1, so we need 101 to run the loop upto 100 
   for b in range(a,101):    # as we have a<=b , we need to start from a 
      if(((1+b**4)%(a*b+1)==0) and (b != a**3) :
      print("a=",a," b=",b)

$(8,30)$ is only such pair.