1. Introduction
Legendary mathematician Pierre de Fermat wrote to his friend on December 25, 1640 that any Pythagorean prime can be expressed as the sum of the squares of two integers [1]. This statement is known as Fermat’s Christmas Theorem or Fermat’s Sum of Two Squares Theorem, one of the most significant propositions in mathematics [2]-[4]. However, it remains largely an open and unsolved problem to find sufficient conditions for the primality of a ( 4 k + 1 )-type of integer to be expressed as a sum of two squares.
Aletheia-Zomlefer [5] states that prime values of quadratic polynomials are conjectural in general. Cox [2] studies to determine which primes can be represented in quadratic forms, indicating that to claim under what conditions a sum of squares will be prime requires analytic conjectures. Iwaniec and Kowalski [6] also emphasize that the general problem of prime values of polynomials, including quadratic cases, remains open.
Pinz [7] indicates that Landau’s problem of n 2 + 1 type is a long-standing challenge. Jacobi [8][9] demonstrates that if an integer n has every ( 4 k + 3 )-type prime factor occurring to an even power, then the number of representations of n as a sum of two squares is 4 ( d 1 ( n ) − d 3 ( n ) ) , where d 1 ( n ) and d 3 ( n ) denote the numbers of ( 4 k + 1 )-type and ( 4 k + 3 )-type divisors of n; however, Jacobi’s theorem does not specify any conditions implying the primality. Jacobi’s formula specifies the number of sum-of-squares representations, which is not the focus of this study.
This brief report sets out when a ( 4 k + 1 )-type integer expressed in a sum of two squares becomes prime. Our finding provides a characterization of primes p ≡ 1 (mod 4) in terms of the uniqueness of sum-of-squares representation. This is a rather elegant result.
2. Preliminaries
Before diving into the main theorem and its proof, we recall some of the well-known results as background knowledge.
Fermat’s Sum of Two Squares Theorem [4] (Fact 1) states that an odd prime p can be written as a sum of two squares if and only if p ≡ 1 (mod 4). Moreover, when such a representation exists, it is unique up to order and signs.
General Representability [4] (Fact 2) states that a positive integer n can be expressed as a sum of two squares if and only if every prime factor of n that is congruent to 3 (mod 4) appears to an even power.
Brahmagupta-Fibonacci Identity [4] (Fact 3) states that the product of two sums of squares is itself a sum of squares, which may appear in two different expressions:
(
x
2
+
y
2
)
(
u
2
+
v
2
)
=
(
x
u
−
y
v
)
2
+
(
x
v
+
y
u
)
2
=
(
x
u
+
y
v
)
2
+
(
x
v
−
y
u
)
2
for real numbers x , y , u , v .
This identity is the engine that generates multiple representations for composite numbers.
3. Main Theorem
Theorem Let a be a square-free positive integer and a ≡ 1 (mod 4). If a = c 2 + d 2 has a unique representation as the sum of two squares (up to order and signs) for a pair of positive integers c and d , then a is prime.
Proof:
First, notice that if a = c 2 + d 2 and g = gcd ( c , d ) > 1 , then g 2 divides a = c 2 + d 2 . But a is square free, so no perfect square greater than 1 can divide it. Therefore, g = gcd ( c , d ) = 1 holds automatically.
We now take a contrapositive proof: if a satisfies all the given conditions but is composite, then a has more than one distinct representation as a sum of two squares.
(Step 1): Let us determine the prime factors of a .
Suppose a satisfies our hypotheses: a ≡ 1 (mod 4), a is square free, and a = c 2 + d 2 ; c , d ∈ N + for some integers with gcd ( c , d ) = 1 . Since a is square free (no prime appears more than once), we have the factorization: a = p 1 p 2 ⋯ p k with each p i being a distinct prime for i = 1 , ⋯ , k .
Let us check what kinds of primes can divide a .
(i) It is easy to see that no prime p i ≡ 0 (mod 4) or p i ≡ 2 (mod 4) can divide a . (ii) Since a is expressible as a sum of two squares, Fact 2 tells that any prime p i ≡ 3 (mod 4) dividing a must appear to be an even power. However, a is square-free, therefore no prime p i ≡ 3 (mod 4) can divide a at all. Therefore, every prime factor of a must be congruent to 1 (mod 4). In other words,
a
=
p
1
p
2
⋯
p
k
where each p i is a distinct prime with p i ≡ 1 (mod 4).
(Step 2): Let us count possible sum-of-squares representations of a .
This counting work follows the basics of [2][4][7] easily. By Fermat’s Two-Square Theorem (Fact 1), each prime p i ≡ 1 (mod 4) has a unique representation p i = x i 2 + y i 2 ; x i , y i ∈ N + .
When we multiply two sums of squares using the Brahmagupta-Fibonacci identity (Fact 3), we get a binary choice at each step. Specifically, if we have built up a representation for p 1 ⋯ p j − 1 , say p 1 ⋯ p j − 1 = A 2 + B 2 for some non-zero integers A and B , and incorporate p j = x j 2 + y j 2 , we then have
(
A
2
+
B
2
)
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j
2
+
y
j
2
)
=
(
A
x
j
−
B
y
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2
+
(
A
y
j
+
B
x
j
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2
or
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A
2
+
B
2
)
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j
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+
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j
2
)
=
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A
x
j
+
B
y
j
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+
(
A
y
j
−
B
x
j
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2
These two choices yield different representations.
Starting with p 1 = x 1 2 + y 1 2 (one representation), at each subsequent prime, we double the number of representations. After incorporating all k primes, we have 2 k − 1 distinct representations. (The factor is 2 k − 1 rather than 2 k because we start with one representation and make k − 1 binary choices.) ■
(Step 3): Let us finish the proof of our theorem.
Suppose a satisfies all the hypotheses and has a unique representation as a sum of two squares.
From Step 1, we know a = p 1 ⋯ p k where all p i ≡ 1 (mod 4) are distinct. Moreover, from Step 2, we know the number of distinct representations is 2 k − 1 .
Since a is assumed to have a unique representation, we have
2
k
−
1
=
1
⇒
k
−
1
=
0
⇒
k
=
1
Consequently, a consists of a single prime factor, which means a is prime.