Here's a method that works 'on paper', no calculator needed.
\(43^{p}(\text{mod}247)=17\) means that \(43^{p}\) when divided by 247, leaves a remainder of 17.
That means that for some k, \(43^{p}-247k=17\), or, equivalently, \(43^{p}-17=247k.\)
Since 247 = 13*19, it follows that \(43^{p}-17\) has to be divisible by both 13 and 19.
Working with mod 13,
\(43\equiv4,\:43^{2}\equiv3,\:43^{3}\equiv-1, \:43^{4}\equiv-4,\text{ etc.}\)
The sequence is easy to work out, the last one, for example, can be worked out using the earlier figures,
\(43^{4}=43^{2}.43^{2}\equiv3\times3=9\equiv-4,\)
or,
\(43^{4}=43.43^{3}\equiv 4\times(-1)=-4\equiv-4.\)
The result is interpreted as the remainder when 43^4 is divided by 13 is 9 (or it overshoots by 4).
(43^4 = 3418801 = 262984*13 + 9, or 43^4 = 3418801 = 262985*13 - 4).
Since \(17\equiv4,\) the sequence for \(43^{p}-17\) begins 0, -1, -5, -8, , ... and what we are looking for are zero entries, (which means that there will be a remainder of zero when dividing by 13).
The first zero (coming from 4 - 4) is saying that 43 - 17 = 26 is divisible by 13.
Running the sequence further produces
0, -1, -5, -8, -7, -3, 0, -1, -5, -8, -7, -3, 0, -1, -5, ...
so we have zeros for p = 1, 7, 13, and at intervals of six thereafter.
Now working with mod 19 to determine when 43^p - 17 is divisible by 19, produces the sequence (noting that \(17\equiv-2(\text{mod}19)\)),
7, 8, -6, 0, 11, 9, -1, 6, 3, 7, 8, -6, 0, 11, ...
so we have zeros for p = 4, 13 and at intervals of nine thereafter.
Putting the two p sequences side by side we have
1, 7, 13, 19, 25, 31, 37, ...
and
4, 13, 22, 31, 40, ....
so we have matching zeros (divisibility by both 13 and 19) at p = 13, 31, ...
