Advanced book on Mathematics Olympiad

Number Theory 705

Hence
[(
p− 1
2

)

!

] 2

≡− 1 (modp),

as desired.
To show that the equation has no solution ifp≡ 3 (mod 4), assume that such a
solution exists. Call ita. Using Fermat’s little theorem, we obtain

1 ≡ap−^1 ≡a^2 ·

p− 21

≡(− 1 )

p− 21

=− 1 (modp).

This is impossible. Hence the equation has no solution.

773.Multiplying the obvious congruences

1 ≡−(p− 1 )(modp),

2 ≡−(p− 2 )(modp),

···

n− 1 ≡−(p−n+ 1 )(modp),

we obtain

(n− 1 )!≡(− 1 )n−^1 (p− 1 )(p− 2 )···(p−n+ 1 )(modp).

Multiplying both sides by(p−n)!further gives

(p−n)!(n− 1 )!≡(− 1 )n−^1 (p− 1 )!(modp).

Because by Wilson’s theorem(p− 1 )!≡− 1 (modp), this becomes

(p−n)!(n− 1 )!≡(− 1 )n(modp),

as desired.
(A. Simionov)

774.Because the common difference of the progression is not divisible byp, the numbers
a 1 ,a 2 , ..., aprepresent different residue classes modulop. One of them, sayai,is
divisible byp, and the others give the residues 1, 2 ,...,p−1 in some order. Applying
Wilson’s theorem, we have

a 1 a 2 ···ap

ai

≡(p− 1 )!≡− 1 (modp);

hencea 1 a 2 ···aapi+1 is divisible byp. Sinceaiis divisible byp, we find thata 1 a 2 ···ap+

aiis divisible byp^2 , as desired.
(I. Cucurezeanu)