Suppose then , and clearly is not a multiple of , so little Fermat tells us that but then is a contradiction.

This also shows by easy generalisation that the congruence cannot be solved for any prime

The definition is iff divides .

Here since divides .

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Another definition is that iff both leave the same remainder modulo . But note that the remainder is between and .

Here and have the same remainder modulo , since .

For a more intuitive explanation, let's consider mod 3 for smaller numbers. If we take all the natural numbers mod 3, we get this: n: 0 1 2 3 4 5 6 7 8 9 ... n % 3: 0 1 2 0 1 2 0 1 2 0 ... And you can see that every time you find a 0, a 2 precedes it. So if we were to extend the definition into the negatives, then we could look and see that 0 % 3 = 0, so then -1 % 3 should be 2 to fit the pattern: n: ... -3 -2 -1 0 1 2 3 4 5 ... n % 3: ... 0 1 2 0 1 2 0 1 2 ... A more algebraic way to look at it is that modulus is defined so that n % m = (n + km) % m for arbitrary integer k. For the case where n = -1 and m = 7, then we get that -1 % 7 = (-1 + 7k) % 7. What if we let k = 1? Then (-1 + 7) % 7 = 6 % 7 = 6, therefore -1 % 7 = 6.

Hidden structure can be interesting. In this case it seems unlikely because the it involves $m$, the number of decimal digits of $2^k-1$ which is not random but also not structured. And the relevant statistic is (perhaps) $10^m \mod 7.$ From what you say there are few enough primes found so far that the lack of any $6 \bmod 7$ is not all that suspicious. One might suspect that there are infinitely many primes and $1/9$ of them are $6 \bmod 5$ in the long run. But also that the gaps between them grow exponentially so one might see only a couple of new primes in a lifetime. · Here is a slightly more general setting which might give perspective: A repunit in base $b$ is a number $r_b(k)=\frac{b^k-1}{b-1}$ I.e. one with all $k$ base digits one. They are prime only if is prime. Their factorization (for small ) is an active project since 1925 with some pretty nice mathematics involved. There is a top priority list of most wanted factorizations as well as next highest more wanted. · So one might write the base $b$ repunits $r_b(k+1)$ and $r_b(k)$, in that order, in base $c$, concatenate, and ask about primality. You ask about the case $b=2,c=10$ , but why those two bases and ,also, why not the other order? Why not a zero in the middle? · I was going to say that questions based on a particular base are less interesting mathematically. I guess it is hard to generalize: · Palindromic primes , particularly in base 10 (and even more so ones like 3,313 built from the decimal digits of $\pi$ ) might strike me as pretty recreational, but you can see from the link that they have gotten a certain measure of attention. · I was going to offer that concatenation for the case $b=c$ would feel more mathematical. I quickly realized that it certainly is mathematical but not new! · I took a quick look at these for $2\leq k \leq 1441$ where we might expect to get $0,1 \bmod 7$ 80 times each, $2,4,5,6$$160$ times each and $3 \bmod 7$$640$ times. That happens except it is $79,641$ for $1,3.$ · The primes occur for $k=2,3,4,7,8,12,19,22,36,46,51,67,79,215,359,394,451$ and $1323.$ · Ignoring the last one which is $3 \bmod 7$, We might expect to get $\bmod 7$ counts of around $1,2,8,2,2,2$ for the first $17. $ It comes out $2,4,7,2,2,0$ · That seems well within what might happen by chance and one would expect primes to be increasingly sparse going forward. · Do you have further data? Out of the next $N$ prime values we might expect $2N/17$ to be $6 \bmod 7.$ Do you have enough that the lack of primes of this sort is remarkable? Given how far your search has gone, how much further would you expect to have to go to have a fair chance of finding one?