a) When is pp (prime) a sum of two squares and justify your answer? [Fill in the blank],

b) Recall every primitive Pythagorean triple can be obtained by choosing relatively prime integers st≥1st\ge1 and setting a=st,b=s2−t22,c=s2+t22.a=st,\:\quad b=\frac{s^2-t^2}{2},\:\quad c=\frac{s^2+t^2}{2}. Can c=1479c=1479 be the hypotenuse of a primitive Pythagorean triple? [Fill in the blank],

What about c=1105c=1105 ? [Fill in the blank],

Question

a) When is pp (prime) a sum of two squares and justify your answer?  [Fill in the blank],

b) Recall every primitive Pythagorean triple can be obtained by choosing relatively prime integers s>t≥1s>t\ge1 and setting a=st,b=s2−t22,c=s2+t22.a=st,\:\quad b=\frac{s^2-t^2}{2},\:\quad c=\frac{s^2+t^2}{2}.  Can c=1479c=1479  be the hypotenuse of a primitive Pythagorean triple? [Fill in the blank],

What about c=1105c=1105 ? [Fill in the blank],

BlackTom AI · Accepted Answer

The prompt contains two parts and two specific c-values to consider, all within a fill-in-the-blank format. I’ll treat each sub-question separately and explain the mathematics behind what makes each statement true or false.

Part a) When is p (prime) a sum of two squares and justify your answer?
- A classic result in number theory, Fermat’s theorem on sums of two squares, tells us exactly which primes can be written as a sum of two squares. According to the theorem, a prime p can be expressed as p = x^2 + y^2 with integers x and y if and only if p = 2 or p ≡ 1 (mod 4).
- Why these conditions? If p ≡ 3 (mod 4), then any representation p = x^2 + y^2 would force p to have a form that cannot hold due to modulo 4 arithmetic and properties of quadratic residues. For p = 2, the representation is trivial: 2 = 1^2 + 1^2. For primes p ≡ 1 (mod 4), there exists a representation, which can be proven via Gaussian integers or other number-theoretic methods.
- Therefore, the fill-in-the-blank statement should reflect the precise condition: p is a sum of two squares exactly when p = 2 or p ≡ 1 (mod 4). If your answer omits these conditions or misstates the congruence class, it misses the complete criterion.

Part b) Recall every primitive Pythagorean triple can be obtained by choosing relatively prime integers s > t ≥ 1 with opposite parity and setting a = st, b = (s^2 − t^2)/2, c = (s^2 + t^2)/2. Can c = 1479 be the hypotenuse of a primitive Pythagorean triple?
- A primitive Pythagorean triple has hypotenuse c equal to the sum of two squares: c = s^2 + t^2, with s and t coprime and of opposite parity. Therefore, for c to be the hypotenuse of a primitive triple, c must be representable as a sum of two squares with a representation that can come from coprime s and t of opposite parity.
- The number 1479 factors as 3 × 17 × 29. To decide whether 1479 can be written as a sum of two squares, use the sum-of-two-squares criterion: a positive integer n is a sum of two squares if and only if in the prime factorization of n, every prime congruent to 3 mod 4 appears with even exponent. Here, 3 ≡ 3 (mod 4) appears to the first power (odd exponent). Since a prime factor congruent to 3 mod 4 appears with an odd exponent, 1479 cannot be expressed as a sum of two squares.
- Consequently, 1479 cannot be the hypotenuse c of any (primitive) Pythagorean triple, because such a triple requires c to be a sum of two squares.

What about c = 1105?
- Let’s factor 1105: 1105 = 5 × 13 × 17. All three prime factors are congruent to 1 modulo 4 (since 5 ≡ 1, 13 ≡ 1, 17 ≡ 1 mod 4). A well-known consequence of the sum-of-two-squares theorem is that if all prime factors of n that are ≡ 3 mod 4 have even exponent (and there are none of them here), then n can be expressed as a sum of two squares. In fact, primes ≡ 1 mod 4 contribute factors that can be represented as sums of two squares, and the product of such numbers also has a representation as a sum of two squares.
- Therefore, 1105 can be written as c = s^2 + t^2 for some integers s > t ≥ 1, with s and t coprime and of opposite parity, yielding a primitive Pythagorean triple with c as the hypotenuse. Hence 1105 can indeed serve as the hypotenuse of a primitive triple, provided the corresponding s and t can be arranged to satisfy the standard primitive triple parameterization.

Summary of reasoning for the blanks (without stating a final explicit answer):
- For primes p, the complete criterion is p = 2 or p ≡ 1 (mod 4).
- For c = 1479, check via the sum-of-two-squares test: presence of a prime ≡ 3 (mod 4) to an odd exponent (here, 3^1) rules it out as a sum of two squares, so it cannot be the hypotenuse of a primitive triple.
- For c = 1105, since all prime factors are ≡ 1 (mod 4), the number passes the sum-of-two-squares criterion, implying it can be represented as s^2 + t^2 with appropriate s and t, and thus can be the hypotenuse of a primitive Pythagorean triple.

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