HMMT 二月 2015 · 代数 · 第 10 题
HMMT February 2015 — Algebra — Problem 10
题目详情
- Find all ordered 4-tuples of integers ( a, b, c, d ) (not necessarily distinct) satisfying the following system of equations: 2 2 2 2 a − b − c − d = c − b − 2 2 ab = a − d − 32 2 ac = 28 − a − d 2 ad = b + c + 31 .
解析
- Find all ordered 4-tuples of integers ( a, b, c, d ) (not necessarily distinct) satisfying the following system of equations: 2 2 2 2 a − b − c − d = c − b − 2 2 ab = a − d − 32 2 ac = 28 − a − d 2 ad = b + c + 31 . Answer: (5 , − 3 , 2 , 3) We first give two systematic solutions using standard manipulations and divisibility conditions (with some casework), and then a third solution using quaternionic number theory (not very practical, so mostly for your cultural benefit). Solution 1. Subtract the second equation from the third to get a ( c − b + 1) = 30. Add the second and third to get 2 a ( b + c ) = − 4 − 2 d . Substitute into the fourth to get 31 a − 2 2 a (2 ad − 31) = − 4 − 2 d ⇐⇒ a (31 − 2 ad ) = 2 + d ⇐⇒ d = , 2 2 a + 1 which in particular gives a 6 ≡ 1 (mod 3). Then plugging in a factor of 30 for a gives us the system of equations b + c = 2 ad − 31 and c − b + 1 = 30 /a in b, c . Here, observe that b + c is odd, so c − b + 1 is even. Thus a must be odd (and from earlier a 6 ≡ 1 (mod 3)), so a ∈ {− 1 , ± 3 , 5 , ± 15 } . Manually checking these, we see that the only possibilities we need to check are ( a, d ) = (5 , 3) , ( − 1 , − 11) , ( − 3 , − 5), corresponding to ( b, c ) = ( − 3 , 2) , (11 , − 20) , (5 , − 6). Then check the three candidates against first 2 2 2 2 condition a − b − c − d = c − b − 2 to find our only solution ( a, b, c, d ) = (5 , − 3 , 2 , 3). Solution 2. Here’s an alternative casework solution. From 2 ad = b + c + 31, we have that b + c is 2 2 odd. So, b and c has different parity. Thus, b + c ≡ 1 (mod 4). Plugging this into the first equation, we get that a and d also have the same parity. 2 2 2 2 So, a − b − c − d ≡ − 1 (mod 4). Thus, c − b − 2 ≡ − 1 (mod 4). So, c ≡ b + 1 (mod 4). From taking modulo a in the second and third equation, we have a | d + 32 and a | 28 − d . So, a | 60. Now, if a is even, let a = 2 k and d = 2 m . Plugging this in the second and third equation, we get 2 kc = 14 − k − m and 2 kb = k − m − 16. So, k ( c − b ) = 15 − k . 15 − k 15 We can see that k 6 = 0. Therefore, c − b = = − 1. k k 15 15 But c − b ≡ 1 (mod 4). So, − 1 ≡ 1 (mod 4), or ≡ 2 (mod 4) which leads to a contradiction. k k So, a is odd. And we have a | 60. So, a | 15. This gives us 8 easy possibilities to check... Solution 3. The left hand sides clue us in to the fact that this problem is secretly about quaternions. Indeed, we see that letting z = a + bi + cj + dk gives ( z − i + j ) z = − 2 − 32 i + 28 j + 31 k. 2 2 2 2 Taking norms gives N ( z − i + j ) N ( z ) = 2 +32 +28 +31 = 2773 = 47 · 59. By the triangle inequality, N ( z ) , N ( z − i + j ) aren’t too far apart, so they must be 47 , 59 (in some order). 5 Thus z, z − i + j are Hurwitz primes. We rely on the following foundational lemma in quaternion number theory: 1+ i + j + k 5 For the purposes of quaternion number theory, it’s simpler to work in the the Hurwitz quaternions H = 〈 i, j, k, 〉 , Z 2 which has a left- (or right-) division algorithm, left- (resp. right-) Euclidean algorithm, is a left- (resp. right-) principal ideal domain, etc. There’s no corresponding division algorithms when we’re working with the Lipschitz quaternions, i.e. those with integer coordinates. Algebra Lemma. Let p ∈ Z be an integer prime, and A a Hurwitz quaternion. If p | N ( A ), then the H A + H p (a left ideal, hence principal) has all element norms divisible by p , hence is nontrivial. (So it’s either H p or of the form H P for some Hurwitz prime P .) In our case, it will suffice to apply the lemma for A = − 2 − 32 i + 28 j + 31 k at primes p = 47 and q = 59 ′ ′ to get factorizations (unique up to suitable left/right unit multiplication) A = QP and A = P Q ′ ′ (respectively), with P, P Hurwitz primes of norm p , and Q, Q Hurwitz primes of norm q . Indeed, ′ these factorizations come from H A + H p = H P and H A + H q = H Q . We compute by the Euclidean algorithm: H A + H (47) = H ( − 2 − 32 i + 28 j + 31 k ) + H (47) = H ( − 2 + 15 i − 19 j − 16 k ) + H (47) − 2 + 15 i − 19 j − 16 k = [ H (47 · 18) + H (47)( − 2 − 15 i + 19 j + 16 k )] 47 · 18 − 2 + 15 i − 19 j − 16 k = [ H 18 + H ( − 2 + 3 i + j − 2 k )] 18 − 2 + 15 i − 19 j − 16 k = H ( − 2 + 3 i + j − 2 k ) 18 − 54 − 90 i + 54 j − 36 k = H 18 = H ( − 3 − 5 i + 3 j − 2 k ) . 6 7 Thus there’s a unit ǫ such that P = ǫ ( − 3 − 5 i + 3 j − 2 k ). ′ Similarly, to get P , we compute A H + 47 H = ( − 2 − 32 i + 28 j + 31 k ) H + 47 H = ( − 2 + 15 i − 19 j − 16 k ) H + 47 H − 2 + 15 i − 19 j − 16 k = [(47 · 18) H + 47( − 2 − 15 i + 19 j + 16 k ) H ] 47 · 18 − 2 + 15 i − 19 j − 16 k = [18 H + ( − 2 + 3 i + j − 2 k ) H ] 18 − 2 + 15 i − 19 j − 16 k = ( − 2 + 3 i + j − 2 k ) H 18 − 54 + 18 i + 18 j + 108 k = H 18 = ( − 3 + i + j + 6 k ) H , ′ ′ ′ so there’s a unit ǫ with P = ( − 3 + i + j + 6 k ) ǫ . ′ Finally, we have either z = ǫ ( − 3 − 5 i + 3 j − 2 k ) for some ǫ , or z − i + j = ( − 3 + i + j + 6 k ) ǫ for some ′ ǫ . Checking the 24 + 24 cases (many of which don’t have integer coefficients, and can be ruled out immediately) gives z = iP = 5 − 3 i + 2 j + 3 k as the only possibility. Remark. We have presented the most conceptual proof possible. It’s also possible to directly compute based on the norms only, and do some casework. For example, since 47 ≡ 3 (mod 4), it’s easy to 2 2 2 2 check the only ways to write it as a sum of four squares are ( ± 5) + ( ± 3) + ( ± 3) + ( ± 2) and 2 2 2 2 ( ± 3) + ( ± 1) + ( ± 1) + ( ± 6) . Remark. For a systematic treatment of quaternions (including the number theory used above), one good resource is On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry by John H. Conway and Derek A. Smith. A more focused treatment is the expository paper Factorization of Hurwitz Quaternions by Boyd Coan and Cherng-tiao Perng. For an example of interesting research in this rather exotic area, see the Metacommutation of Hurwitz primes paper by Henry Cohn and Abhinav Kumar. 6 2 2 2 2 Hidden computations: we’ve used 47 · 18 = 846 = 2 + 15 + 19 + 16 , and 18 = N ( − 2 + 3 i + j − 2 k ). ± 1 ± i ± j ± k 7 i.e. one of ± 1 , ± i, ± j, ± k, 2 Algebra