COMP 547: Assignment 1 Solutions

Combining these two directions proves that f(a) ∈ QR p if and only
if a does not give the key to √ x.
Lemma 2: f is a one-to-one function.
Proof: Suppose, for the sake of deriving a contradiction, that there exist
two elements i, j ∈ {1, 2, . . . , p − 1}\{ √ x, p − √ x}, i ≠ j, such that
f(i) = f(j) = g. Then, by the definition of f, (i − √ x)g ≡ i +
√ x (mod p) and (j −
√ x)g ≡ j +
√ x (mod p). Combining these
two results, we obtain the following sequence of statements: (NB:
Here, we use some multiplicative cancellation. This is valid, since
gcd(p, x) = 1 for any integer x, given that p is prime. We also know
that it is not the case that i− √ x ≡ 0 (mod p) or j− √ x ≡ 0 (mod p),
since i, j /∈ { √ x, p − √ x} and it is not the case that g ≡ 0 (mod p),
since f’s range is {2, 3, . . . , p − 1}.)
(i − √ x)(j + √ x)g ≡ (i + √ x)(j − √ x)g (mod p)
(i − √ x)(j + √ x) ≡ (i + √ x)(j − √ x) (mod p)
ij + i √ x − j √ x − x ≡ ij − i √ x + j √ x − x (mod p)
i √ x − j √ x ≡ −i √ x + j √ x (mod p)
√ x(i − j) ≡
√ x(j − i) (mod p)
i − j ≡ j − i (mod p)
2i ≡ 2j (mod p)
i ≡ j (mod p)
Since i, j ∈ {1, 2, . . . , p − 1}, this last result can occur only if i = j.
This contradicts our assumption that i ≠ j. Therefore, we conclude
that there cannot exist two elements i, j ∈ {1, 2, . . . , p−1}, i ≠ j, such
that f(i) = f(j). This establishes that f is a one-to-one function.
Lemma 3: If p ≡ 1 (mod 4), then p − 1 ∈ QR p .
Proof: We know that x ∈ QR p if and only if x p−1
2 ≡ 1 (mod p). Let x =
p−1. Then p−1 ∈ QR p if and only if (p−1) p−1
2 ≡ 1 (mod p). Since
p ≡ 1 (mod 4), we know that 4|p−1, so 4k = p−1 for some integer k.
Then p − 1 ∈ QR p if and only if (4k) 2k ≡ 1 (mod 4k + 1). Note that
for any integer b, (b + 1)(b − 1) = b 2 − 1. So (b + 1)|(b 2 − 1). So b 2 ≡
1 (mod b + 1). Thus, (4k) 2 ≡ 1 (mod 4k + 1), and it immediately
follows that (4k) 2k ≡ 1 (mod 4k + 1). This was our condition for
p − 1 being in QR p . Thus, we can conclude that p − 1 ∈ QR p .
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