# Characterization of \( n = a^2 + ab + b^2 \) ## General and Properties The form \( n = a^2 + ab + b^2 \) is known as a **quadratic form**, specifically the norm form for the Eisenstein integers (\( \mathbb{Z}[\omega] \), where \( \omega = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \)), since \[ |a + b\omega|^2 = (a + b\omega)(a + b\overline{\omega}) = a^2 + ab + b^2. \] ## Characterization A positive integer \( n \) can be written as \( n = a^2 + ab + b^2 \) with \( a, b \in \mathbb{Z} \) **if and only if** in the prime factorization of \( n \), every prime congruent to \( 2 \bmod 3 \) occurs to an even power. ### Details - **Prime factorization:** \( n = 3^{e_} \prod_{p \equiv 1 \bmod 3} p^{e_p} \prod_{q \equiv 2 \bmod 3} q^{f_q} \) - **Condition:** All \( f_q \) are even (i.e., primes \( q \equiv 2 \pmod{3} \) appear to even exponents) This is analogous to the classic result that an integer is a sum of two squares iff all primes \( \equiv 3 \bmod 4 \) appear with even exponent. ## Why? - **Primes \( p = 3k+1 \) or \( p=3 \):** Can be represented as \( a^2+ab+b^2 \). - **Primes \( p = 3k+2 \):** Cannot be represented, but their squares can. --- # Test Case: \( n = 2025 \) ## Step 1: Factorize 2025 \[ 2025 = 5^2 \times 9^2 = (5 \times 9)^2 = 45^2 \] But for prime factorization: \[ 2025 = 5^2 \times 3^4 \] - 3: \( 3 \equiv \bmod 3 \) - 5: \( 5 \equiv 2 \bmod 3 \) The exponent of 5 (a \( 2 \bmod 3 \) prime) is **even** (2). The exponent of 3 is 4. Thus, **2025 is representable** in the given form. --- ## Step 2: Find All Integer Pairs \( (a, b) \) with \( a^2 + ab + b^2 = 2025 \) We seek integer solutions to \[ a^2 + ab + b^2 = 2025 \] ### Symmetries - Swapping \( a \) and \( b \) does not change the value. - Changing sign of both \( a \) and \( b \) does not change the value. Thus, it suffices to find unordered pairs up to sign. ### Approach For fixed \( a \), solve the quadratic in \( b \): \[ b^2 + ab + (a^2 - 2025) = \] The discriminant must be a perfect square: \[ \Delta = a^2 - 4(a^2 - 2025) = -3a^2 + 810 \] So \( -3a^2 + 810 = k^2 \) for integer \( k \). \[ 3a^2 + k^2 = 810 \] Now, \( 3a^2 \leq 810 \implies |a| \leq 52 \) For each integer \( a \) from \(-52\) to \(52\), check if \( 810 - 3a^2 \) is a perfect square. #### Let's Solve for All Integer \( a \): Let \( k^2 = 810 - 3a^2 \). So \( k \geq \), \( 810 - 3a^2 \geq \). \[ 3a^2 \leq 810 \implies a^2 \leq 270 \implies |a| \leq 51 \] For \( a = -51 \) to 51: For each \( a \), compute \( m = 810 - 3a^2 \). If \( m \) is a perfect square, then set \( k = \sqrt{m} \). #### Let's Find All Such \( a \): 1. \( a = : \quad k^2 = 810 \implies k = 90 \) 2. \( a = \pm 30: \quad 3a^2 = 270 \implies k^2 = 540 \) (not a perfect square) 3. \( a = \pm 36: \quad 3a^2 = 3888 \implies k^2 = 4212 \) (not a perfect square) 4. \( a = \pm 45: \quad 3a^2 = 6075 \implies k^2 = 2025 \implies k = 45 \) 5. \( a = \pm 51: \quad 3a^2 = 7803 \implies k^2 = 297 \) (not a perfect square) Let's systematically try all \( a \) for which \( m = 810 - 3a^2 \) is a perfect square. #### Let's look for all \( a \) where \( 810 - 3a^2 \) is a perfect square: Let \( k \geq \), \( 3a^2 + k^2 = 810 \). We can also try small possible values of \( a \): - \( a = \Rightarrow k^2 = 810 \rightarrow k = 90 \) - \( a = \pm 45 \Rightarrow 3 \times 2025 = 6075 \rightarrow k^2 = 810 - 6075 = 2025, k = 45 \) - \( a = \pm 51 \Rightarrow 3 \times 2601 = 7803 \rightarrow k^2 = 810 - 7803 = 297 \) (not a square) - \( a = \pm 30 \Rightarrow 3 \times 900 = 270 \rightarrow k^2 = 810 - 270 = 540 \) (not a square) - \( a = \pm 42 \Rightarrow 3 \times 1764 = 5292 \rightarrow k^2 = 810 - 5292 = 2808 \) (not a square) - \( a = \pm 39 \Rightarrow 3 \times 1521 = 4563 \rightarrow k^2 = 810 - 4563 = 3537 \) (not a square) - \( a = \pm 18 \Rightarrow 3 \times 324 = 972 \rightarrow k^2 = 810 - 972 = 7128 \) (not a square) - \( a = \pm 9 \Rightarrow 3 \times 81 = 243 \rightarrow k^2 = 810 - 243 = 7857 \) (not a square) Now let's try \( a = \pm 90 \Rightarrow 3 \times 810 = 24300 \), which is greater than 810, so not allowed. Alternatively, let’s try to find all integer solutions to \( 3a^2 + k^2 = 810 \): Let’s try for \( k = \) upwards: - \( k = \implies a^2 = 270 \implies a \) not integer. - \( k = 9 \implies k^2 = 81 \implies 3a^2 = 8019 \implies a^2 = 2673 \) not integer. - \( k = 18 \implies 324 \implies 3a^2 = 7776 \implies a^2 = 2592 \) not integer. - \( k = 45 \implies k^2 = 2025 \implies 3a^2 = 6075 \implies a^2 = 2025 \implies a = \pm 45 \) - \( k = 63 \implies k^2 = 3969 \implies 3a^2 = 4131 \implies a^2 = 1377 \) not integer. - \( k = 72 \implies k^2 = 5184 \implies 3a^2 = 2916 \implies a^2 = 972 \) not integer. - \( k = 81 \implies k^2 = 6561 \implies 3a^2 = 1539 \implies a^2 = 513 \) not integer. - \( k = 90 \implies k^2 = 810 \implies 3a^2 = \implies a = \) So, the only integer values for \( a \) are \( a = , \pm 45 \). #### Now, For Each \( a \), Find \( b \): Recall: \[ b^2 + ab + a^2 - 2025 = \] So, for each \( a \), \( b \) is a root of: \[ b^2 + ab + (a^2 - 2025) = \] \[ b = \frac{-a \pm \sqrt{a^2 - 4(a^2 - 2025)}}{2} = \frac{-a \pm \sqrt{-3a^2 + 810}}{2} \] ### Case 1: \( a = \) \[ b^2 + () b + ( - 2025) = \implies b^2 = 2025 \implies b = \pm 45 \] ### Case 2: \( a = 45 \) \[ b^2 + 45b + (2025 - 2025) = \implies b^2 + 45b = \implies b(b + 45) = \implies b = , b = -45 \] ### Case 3: \( a = -45 \) \[ b^2 -45b + (2025 - 2025) = \implies b^2 - 45b = \implies b = , b = 45 \] --- # List of Integer Pairs \( (a, b) \) (up to order and sign) Collecting all, and factoring out symmetry (swapping and negation): - \( (, 45) \) - \( (45, ) \) - \( (, -45) \) - \( (-45, ) \) - \( (45, -45) \) - \( (-45, 45) \) But under swapping and negation, these collapse to two distinct types: - \( (, 45) \) (and its equivalents) - \( (45, -45) \) (and its equivalents) ## Final List (up to order and sign) \[ (a, b) = (, 45),\quad (45, -45) \] --- ## Summary ### 1. **Characterization:** A positive integer \( n \) can be written as \( n = a^2 + ab + b^2 \) with \( a, b \in \mathbb{Z} \) **if and only if** all primes \( \equiv 2 \pmod{3} \) in \( n \)'s factorization appear to even powers. ### 2. **For \( n = 2025 \):** All integer pairs \( (a,b) \) (up to order and sign) are: - \( (, 45) \) - \( (45, -45) \)  *Eisenstein integer lattice; points at squared distance 2025 correspond to solutions.*