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Let $p$, $q$ be two primes such that $q \equiv 2 \pmod{5}$ and $p = 4q+1$.

Show that $$100^{q} \not\equiv 1\pmod{p} $$

Here is one way that I tried to tackle this (and failed, obviously...):

Assume by contradiction that $100^{q} \equiv 1\pmod{p} $. By Fermat's little Theorem it follows that: $$100^{q} \equiv 100 \pmod{q}$$ This is where I got stuck. However, I did notice that:
$$p \equiv 1\pmod{q}$$ $$p \equiv -1\pmod{5}$$

But these congruences didn't help me either.

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If we negate the conclusion, we have $$ 100^q\equiv 1 (\textrm{ mod }p),$$ and this is equivalent to the existence $x$ in the following $$ 100\equiv x^4 (\textrm{ mod }p)$$ This reduces to $ 10 \equiv x^2 (\textrm{ mod }p)$, or $ -10 \equiv x^2 (\textrm{ mod }p)$

Both of them cannot be true since $$ \left(\frac{10}{p}\right)=\left(\frac{-10}{p}\right) = -1.$$

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  • $\begingroup$ Wow, what a great answer ! Thanks ! $\endgroup$ – Robert777 May 30 '13 at 23:27
  • $\begingroup$ why $100\equiv x^4 (\textrm{ mod }p)$? $\endgroup$ – Lrrr May 31 '13 at 6:45
  • $\begingroup$ books.google.co.il/… Theorem 4.13 $\endgroup$ – Robert777 May 31 '13 at 7:39

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