Determine the least positive residues $x$ in the following cases (1987 is a prime):
(a) $25^{995} \equiv x \pmod {1987}$
(b) $26^{995} \equiv x \pmod {1987}$
(a) Since $5^2=25$, then 25 is a quadratic residue of 1987. By Euler's Criterion this means
$$ 25^{\frac{1987-1}{2}} \equiv 25^{993} \equiv 1 \pmod {1987} $$
Multiplying through by $25^2=625$ gives
$$ 25^{995} \equiv 625 \pmod {1987} $$
And so the least positive residue $x$ in $25^{995} \equiv x \pmod {1987}$ is $x=625$.
(b) We test whether 26 is a quadratic residue of 1987.
$$ \begin{align} (\frac{26}{1987}) & = (\frac{2}{1987}) \times (\frac{13}{1987}) \\ \\ & = (-1) \times (\frac{13}{1987}) \\ \\ & = (-1) \times (\frac{1987}{13}) \\ \\ & = (-1) \times (\frac{11}{13}) \\ \\ & = (-1) \times (\frac{13}{11}) \\ \\ & = (-1) \times (\frac{2}{11}) \\ \\ & = (-1) \times (-1) = 1 \end{align} $$
This means 26 is a quadratic residue of 1987. And so by Euler's Criterion
$$ 26^{\frac{1987-1}{2}} \equiv 26^{993} \equiv 1 \pmod {1987} $$
Multiplying through by $26^2=676$ we have
$$ 26^{995} \equiv 676 \pmod {1987} $$
And so the least positive residue $x$ in $26^{995} \equiv x \pmod {1987}$ is $x=676$.