In which finite fields Zp does -1 have a square root?

The answer is p = 2 /in Z2 we have -1=1, i.e. 1² = -1/

and all odd p ≡ 1 (mod 4); -1 is not perfect square in Zp if

p ≡ 3 (mod 4). This is a very well-known theorem that can be found in almost every book on numbers theory. Here is a proof. Note that (p-1)/2 is even exactly for all p ≡ 1 (mod 4) and is odd exactly for all p ≡ 3 (mod 4). According the Fermat's Little Theorem:

(F) x^(p-1) ≡ 1 (mod p) for x = 1,2,3, . . , p-1 - all they are co-prime with p /they are all non-zero elements of Zp/. Suppose -1 is a square of some x:

x² = -1 in Zp, that means x² ≡ p-1 ≡ -1 (mod p), then /see (F) above/

x^(p-1) ≡ (x²)^((p-1)/2) ≡ (-1)^((p-1)/2) ≡ 1 (mod p), so (p-1)/2 must be even, or p ≡ 1 (mod 4)..

Now let p ≡ 1 (mod 4), then (-1)^((p-1)/2) ≡ 1 (mod p),

(p-1)/2 being even, applying (F) we obtain that all numbers

(S) 1², 2², 3², . . , ((p-1)/2)² are solutions of the congruence

(C) x^((p-1)/2) ≡ 1 (mod p)

They are all different in Zp, otherwise if

x² ≡ y² (mod p) we would have (x - y)(x + y) ≡ 0 (mod p), the latter impossible having x - y and x + y less than p, hence co-prime with p. According Lagranges' Theorem (C) cannot have more solutions, than its degree (p-1)/2, so the above numbers (S) are all its solutions. But -1 is also solution of (C), so -1 is one of the numbers in (S), the latter means x² = -1 for some x, or, say it otherwise, -1 is a square.

Clearly... for all primes p that can be written as n^2 + 1 for some integer n.

then -1 has a square root. it is congruent to n (modulo p)

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a million. sq. root 8x-a million (end of sq. root) = 5 8x -a million = 25 (squarethe two factors) 8x = 26 x = 13 /4 --answer 2. sq. root x-7 (end of sq. root) +5 = 11 squareroot x-7 (end of sq. root) = 6 x- 7 = 36 x = 40 3 --answer