HMMT 二月 2015 · 代数 · 第 10 题

英文解析

10. Find all ordered 4-tuples of integers ( a, b, c, d ) (not necessarily distinct) satisfying the following systemof 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 anddivisibility conditions (with some casework), and then a third solution using quaternionic numbertheory (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 secondand 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 a + 12
    which in particular gives a 6 ≡ 1 (mod 3). Then plugging in a factor of 30 for a gives us the system ofequations 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 } . Manuallychecking 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).
    Thus z, z − i + j are Hurwitz primes. We rely on the following foundational lemma in quaternion 5
    number theory:
    1+ i + j + k
    For the purposes of quaternion number theory, it’s simpler to work in the the Hurwitz quaternions H = 〈 i, j, k, 〉 ,5
    2Zwhich has a left- (or right-) division algorithm, left- (resp. right-) Euclidean algorithm, is a left- (resp. right-) principal idealdomain, etc. There’s no corresponding division algorithms when we’re working with the Lipschitz quaternions, i.e. those withinteger 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 )] − 2 + 15 i − 19 j − 16 k 18 = H ( − 2 + 3 i + j − 2 k ) − 54 − 90 i + 54 j − 36 k 18 = H = H ( − 3 − 5 i + 3 j − 2 k ) .18
    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 ] − 2 + 15 i − 19 j − 16 k 18 = ( − 2 + 3 i + j − 2 k ) H − 54 + 18 i + 18 j + 108 k 18 = H = ( − 3 + i + j + 6 k ) H ,18
    ′ ′ ′
    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 outimmediately) gives z = i p = 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 computebased 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), onegood 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 Hurwitzprimes 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 ± ki.e. one of ± 1 , ± i, ± j, ± k,7
    Algebra 2