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If $p$ is a prime, $(a,p)=1$,denote $ord(a,p)=d,$ where $d$ is the smallest positive integer solution to the equation $a^d\equiv 1 \pmod p$.We can prove that $$10^n\equiv -1 \pmod p\tag1$$ is solvable iff $ord(10,p)$ is even.

Now,consider this equation,$$10^a+10^b\equiv -1 \pmod p.\tag2$$ If $10$ is a primitive root modulo $p$, then there is an integer $a$ for every $b≠\dfrac{p-1}{2}\pmod {p-1}$ so that $a,b$ satisfies $(2)$.

My question is,what's the necessary and sufficient condition that $(2)$ has at least $1$ solution?

If we are given a prime $p$,how to determine whether $(2)$ is solvabe?

This is a way,but not effectively: for every positive integer $b\leq\frac{1}{2}ord(10,p)$,determine whether $(2)$ is solvabe for $a$.By this way,I find $(2)$ is solvable for these primes,which $10^n\equiv -1 \pmod p$ is not solvable:

$3,31,37,43,53,67,71,83,107,151,163,173,191,199,227,277,283,307,311,317,347,359,397,431,439,443,467,479,523,547,563,587,599,613,631,643,683,719,751,757,773,787,797,827,839,853,883,907,911,919,947,991,\cdots$

My original problem is: how many "$1$" is need at least for a decimal number which is consisting of "$0$" and "$1$" and divisible by $p$?This question is to find these primes that three "$1$" is need at least.

Thank you.

lsr314
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2 Answers2

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The first few primes for which it is not solvable are $$5, 11, 41, 73, 79, 101, 137, 239, 271, 281, 641, 733, 859, 1321, 1409, 2531, 2791, 3191, 3541, 4013, 4093, 4637, 4649, 5051, 5171, 5237, 6163, 6299, 7253, 7841, 8779, 9091, 9161, 11831, 12517, 12671$$

The sequence does not seem to be in the OEIS.

On the other hand, if you replace $10$ by $2$ there is https://oeis.org/A179113

EDIT: I doubt that there's a simple necessary and sufficient condition. But here might be the start of a heuristic analysis which might suggest there should be infinitely many.

Heuristically, if the order of $10$ mod $p$ is $m$, we evaluate $10^a + 10^b \mod p$ for approximately $m^2/2$ unordered pairs $(a,b)$, so the probability that none of those is congruent to $-1 \mod p$ should be approximately $\exp(-m^2/(2p))$. So if you want to find primes $p$ for which your equation is not solvable, you might look at those where the order of $10 \mod p$ is less than about $\sqrt{p}$.

The order of $10 \mod p$ is $m$ or one of its divisors if $p$ divides $10^m - 1$. So we might look for primes $ p > m^2$ dividing $10^m - 1$. Nearly all positive integers $x$ will have at least one prime factor greater than $\log_{10}(x)^2$ (see Dickman's theorem).

Robert Israel
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The thing looks pretty general. Consider 7 in base 2. Here 2+4-1, mod 7, even though 2 is not a primitive root. The thing is resolving to what conditions are needed for p to divide the sum of three powers of 10. The number 37 divides any such sum as long as the three powers are different, modulo 3, eg, 100,011.

Having a period that is divisible by $3$ implies a solution must exist, for this reason.

wendy.krieger
  • 6,856
  • My question is to find all these primes,i gave a necessary and sufficient condition so that $(1)$ is solvable,what about $(2)$? – lsr314 May 19 '13 at 06:16