more numb’-th’y exercisen

let p be an odd prime &
let g & g’ be primitive
roots (mod p).

any Primitive Root, h, satisfies
(h^{{p-1}\over2})^2 = 1 (mod p);
since h^{{p-1}\over2}\not= 1 (mod p)
[this uses “h is primitive”],
we can conclude that
h^{{p-1}\over2} ~ -1 (mod p).

etcetera. forget it.
the handwritten stuff
is beautiful though.
\TeX is too hard.
it’s not so bad in the real editor,
of course.

oh, ps. (gg’)^[(p-1)/2]
is now seen to be congruent
to 1… and so is not a
primitive root (which, on
the day, was to’ve been shown).


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