Questions: Sum of Two Squares Theorem

Question 1

The number 45 = 3² × 5. Can 45 be expressed as a sum of two squares?

• A. No — because 3 is a prime of the form 4k+3, it blocks all representations
• B. Yes — because 3 appears to an even power (2), it does not block representation
• C. No — because 5 ≡ 1 (mod 4) primes can never be sums of two squares
• D. Yes — but only because 45 is composite, not prime

Answer: (B)

The theorem states that 4k+3 primes must appear to *even* powers. In 45 = 3² × 5, the only 4k+3 prime is 3, and it appears to the power 2 (even). So 45 is representable: 45 = 3² + 6² = 9 + 36. The common misconception (option A) is treating any appearance of a 4k+3 prime as a total block — but only *odd-power* appearances block representation.

Question 2

Which of the following integers CANNOT be expressed as a sum of two squares?

• A. 25 = 5²
• B. 50 = 2 × 5²
• C. 63 = 3² × 7
• D. 65 = 5 × 13

Answer: (C)

63 = 3² × 7. The prime 7 ≡ 3 (mod 4) appears to the first power (odd), so 63 cannot be expressed as a sum of two squares. Check the others: 25 = 5² (5 ≡ 1 mod 4, works: 3² + 4²), 50 = 2 × 5² (both 2 and 5 are not 4k+3 primes, works: 5² + 5²), 65 = 5 × 13 (both ≡ 1 mod 4, works: 4² + 7²). Only 63 has a 4k+3 prime appearing to an odd power.

Question 3

The prime 7 cannot be expressed as a sum of two squares because 7 ≡ 3 (mod 4).

• T. True
• F. False

Answer: True

True. The theorem states that a prime p is expressible as a² + b² if and only if p = 2 or p ≡ 1 (mod 4). Since 7 ≡ 3 (mod 4), it belongs to the 4k+3 class, which is never a sum of two squares. You can verify directly: the only pairs of non-negative integers with a² + b² ≤ 7 are (0,1), (0,2), (1,1), (1,2) — giving squares 1, 4, 2, 5. None sum to 7.

Question 4

The integer 9 = 3² cannot be expressed as a sum of two squares because 3 is a prime of the form 4k+3.

• T. True
• F. False

Answer: False

False. 9 = 3² + 0². While 3 ≡ 3 (mod 4), it appears to the *second* power (even) in 9 = 3². The theorem requires that 4k+3 primes appear to even powers — and a prime to an even power 2k equals p^(2k) = (p^k)² + 0². So 9 is representable: 9 = 3² + 0². Only an *odd* power of a 4k+3 prime blocks representation.

Question 5

Why does a prime p ≡ 3 (mod 4) appearing to an odd power in the factorization of n block n from being a sum of two squares?

Model answer: In the Gaussian integers ℤ[i], a prime p ≡ 3 (mod 4) remains prime (stays inert) — it does not factor as a product of conjugates π·π̄. This means p cannot be written as |π|² = a² + b². When p appears to an odd power, the Gaussian integer factorization of n contains an irreducible factor that cannot be paired with its conjugate. Brahmagupta's identity shows that products of sums of two squares remain sums of two squares, but an unpaired 4k+3 prime breaks this: it contributes a factor that cannot be absorbed into any (a² + b²) representation.

The proof via Gaussian integers is the cleanest way to see this. Primes ≡ 1 (mod 4) factor as p = π·π̄ in ℤ[i], so p = |π|² = a² + b². Primes ≡ 3 (mod 4) remain inert (prime in ℤ[i]) and cannot be decomposed this way. An even power p² = (p + 0i)(p − 0i) = p² + 0² can still be represented (trivially). An odd power introduces an irreducible factor without a conjugate match, making representation impossible.