Suppose that $n$ is such number and: $n = p_1^{\alpha_1}p_2^{\alpha_2}...p_m^{\alpha_m}$ so for $i=1,2,...,m$ we have: $x^{25} \equiv x \mod{p_i^{\alpha_i}}$ By plugging $x=p_i$ we can easily see that $\alpha_i=1$. Now, for every prime $p_i$ there exist a primitive root mod $p_i$. Call it $g_i$. Since order of $g_i$ is $p_i-1$ and we have $x^{24} \equiv 1 \mod{p_i}$ it follows that $p_i-1|24$. Now it's only a matter of simple calculating.

Isnt this Bulgaria 1995 Mathematical Olympiad ?

It could be, I don't have access to these sources. Can anyone confirm this? Thanks for the report by the way.

Yes it is confirmed . See this ......this is what I could get from here http://www.math.bas.bg/bcmi/

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