Is $n^7 - 77$ ever a Fibonacci number?

Solution 1:

The fibonacci sequence goes as follows:

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} F_n &1&1&2&3&5&8&13&21&34&55&89&144&233&377&610&987\\ F_n\pmod{29}&1&1&2&3&5&8&13&21&5 &26&2&28&1&0&1&1\end{array}$

It is clear that the cycle will repeat for $F_n\pmod{29}$ as we reached again a point where it is two ones in a row.

So, the set of possible values modulo $29$ that a fibonacci number can be is $\{0,1,2,3,5,8,13,21,26,28\}$

On the other hand, $n^7\pmod{29}$ can only ever be $0,1,12,17,28$ by fermat's little theorem.

Thus, $n^7-77$ can only ever be one of $\{9,10,11,22,27\}$

We see that the set of possible values of $n^7-77\pmod{29}$ does not intersect the set of possible values of $F_n\pmod{29}$, therefore no solutions exist.