Determine the least positive residues $x$ in the following cases (1993 is prime):
(a) $25^{997} \equiv x \pmod {1993}$
(b) $26^{997} \equiv x \pmod {1993}$
(a) We observe that $5^2 \equiv 25 \pmod {1993}$, and $1993 \not \mid 25$, and so by Euler's Criterion (7.5)
$$ 25^{\frac{1993-1}{2}} \equiv 25^{996} \equiv 1 \pmod {1993}$$
Multiplying through by 25 gives us
$$ 25^{997} \equiv 25 \pmod {1993}$$
And so the least positive residue is $x=25$.
(b) We try to apply the same technique as in part (a). To do this we need to establish whether 26 is a quadratic residue of 1993.
We proceed
$$ \begin{align} (\frac{26}{1993}) & = (\frac{2}{1993}) \times (\frac{26}{1993}) \\ \\ &= 1 \times (\frac{26}{1993}) \tag{Proposition 7.15} \\ \\ & = (\frac{1993}{13}) = (\frac{2^2}{13}) = 1 \tag{Corollary 7.17} \end{align} $$
And so 26 is a quadratic residue of 1993, and so by Euler's Criterion
$$ 26^{\frac{1993-1}{2}} \equiv 26^{996} \equiv 1 \pmod {1993}$$
Multiplying through by 26 gives us
$$ 26^{997} \equiv 26 \pmod {1993}$$
And so the least positive residue is $x=26$.