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Given that i have: $$10^{a} \equiv 10^{b} \pmod p$$ and we know that: $$a > b$$

Can we say that b is a multiple of a or this is not valid?


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For $p = 2,3,5$ we have $10^a \equiv 10^b \pmod p$ for all $a$, $b$. – martini Nov 1 '12 at 18:06
$10^3-10^2$ must be divisible by some prime, $p$, but $2\not\mid 3$ – Thomas Andrews Nov 1 '12 at 18:11

No, e.g. $\rm\: 10^n\equiv 1\:\Rightarrow\:10^{2n}\equiv 1 \equiv 10^{3n}\:$ but $\rm\:2n\nmid 3n,\:\ 3n\nmid 2n.\:$ However, it is true that if $\rm\:\ell\:$ is the order of $\rm\,10\pmod p,\:$ i.e. the least positive $\rm\:n\:$ such that $\rm\:10^{n}\equiv 1\pmod p,\:$ then it follows that $\rm\:10^k\equiv 1\:\Rightarrow\: \ell\mid k.\:$ For a proof see this answer.

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If $p\mid (10^b-10^a)\implies p\mid 10^a(10^{b-a}-1)$

Clearly, $10^a$ is divisible by $2$ and $5 \iff a\ge 1$

For the other primes $(10,p)=1,$

if $ord_p10=d$ and $10^x\equiv 1\pmod p$ where $x>0$, we know $d$ must divide $x$ which is necessary as well as sufficient condition.

So, $d\mid(b-a)\iff 10^{b-a}\equiv 1\pmod p$

$a,b$ need to maintain the condition, no multiplicity is required.

Special case :

$\space\space\space$if $d\mid a,d\mid b$,

$\space\space\space\space\space\space$ if $d=a, d\mid b\implies a\mid b$

The maximum value of $d$ can be $\phi(p)=p-1,$

so if $(p-1)\mid(b-a), 10^{b-a}\equiv 1\pmod p\implies p\mid (10^b-10^a)$

This can be established using Fermat's Little Theorem,too.

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