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I just do not have a clue how to solve this exercise:

Find every natural number $m$ such that $998^m-1$ divides number $1994^m$.

A friend of mine solved it this way:

$998^m - 1 = (998 - 1) \cdot (1 + 998 + \cdots + 998^{m - 1}) = 997 \cdot (1 + 998 + \cdots + 998^{m - 1})$,

So $997$ is a divisor of $988^m - 1$

And $997$ is a divisor of $1994^m$

Is his solution correct?

Thank you for your time.

iam_agf
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martina
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1 Answers1

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First notice that $1994=2\cdot 997$. It must happen that if $998^m-1$ divides $1994^m=997^m\cdot 2^m$, then

$$998^m-1= 997^k*2^q$$

for $k,q\leq m$. See that $998^m-1$ is always odd, so it won't have factors $2$. So

$$998^m-1=997^k$$

Then, if $m>1$, look that by the Zsigmondy theorem, it must happen that exists a prime $p$ such that $p\mid 998^m-1$, but doesn't divide $998^1-1$, so $998^m-1$ can't be a power of $997$, because it has another prime factor different than $997$. So the only answer will be $m=1$.

iam_agf
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  • Thank you very much. it is quite clear now, but I do not understand, how exactly have you applied Zsigmondy theorem. I understand Zsigmondy theorem like this:For every $n>=1$, there is a prime number $p$, that devides $a^{n}-b^{n}$ . and there is a $k<n$ for that $ p$ does`t devide $a^{k}-b^{k}$ . And I just do not see, how have you applied this in your solution. – martina Nov 18 '16 at 23:50
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    Well, is very easy: if I have a prime factor $ p $ that divides $t_m=998^m-1$ but doesn't divide $998-1=997$, then $ t_m $ can't be a power of $997$. – iam_agf Nov 19 '16 at 00:25