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USA Mathematical Talent Search Solutions To Problem 3/4/16

The document summarizes solutions to a problem posed by the USA Mathematical Talent Search to find a polynomial function with integer coefficients such that the sign of the function is the same as the sign of a related expression for all integer values of the variables. Two example solutions are provided and edited by Richard Rusczyk. The first solution uses a lemma to show that the polynomial a3 + 2b3 + 4c3 - 6abc has the desired properties. The second solution arrives at the same polynomial by setting the related expression equal to zero and manipulating the equations.

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68 views2 pages

USA Mathematical Talent Search Solutions To Problem 3/4/16

The document summarizes solutions to a problem posed by the USA Mathematical Talent Search to find a polynomial function with integer coefficients such that the sign of the function is the same as the sign of a related expression for all integer values of the variables. Two example solutions are provided and edited by Richard Rusczyk. The first solution uses a lemma to show that the polynomial a3 + 2b3 + 4c3 - 6abc has the desired properties. The second solution arrives at the same polynomial by setting the related expression equal to zero and manipulating the equations.

Uploaded by

Arsy
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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USA Mathematical Talent Search

Solutions to Problem 3/4/16


www.usamts.org

3/4/16. Find, with proof, a polynomial f (x, y, z) in three variables, with integer coefficients, such that for all integers a, b,
c, the
sign of f (a, b, c) (that is, positive, negative, or
zero) is the same as the sign of a + b 3 2 + c 3 4.
Credit This problem was devised by Dr. Erin Schram of the National Security Agency.
Comments Most students used algebraic manipulation to arrive at a solution; Tony Liu
and Laura Starkston give example solutions.
Solutions edited by Richard Rusczyk.
Solution 1 by: Tony Liu (10/IL)
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We claim that the polynomial f (a, b, c) = a3 + 2b3 + 4c3 6abc has the desired properties.
Our proof begins with the following lemma:
Lemma: The expressions s = p3 + q 3 + r3 3pqr and t = p + q + r have the same sign for
real numbers p, q, r that are not all equal.
Proof: We note the following identity:
p3 + q 3 + r3 3pqr = (p + q + r)(p2 + q 2 + r2 pq qr rp)
1
(p + q + r)((p q)2 + (q r)2 + (r p)2 ),
=
2
or equivalently,
t
s = ((p q)2 + (q r)2 + (r p)2 ).
2
Note that (p q)2 + (q r)2 + (r p)2 0, with equality if and only if p = q = r. By hypothesis, this cannot hold, so (pq)2 +(q r)2 +(r p)2 > 0. Thus, t = 0 if and only if s = 0.
Moreover, when s, t 6= 0, we may divide by t to get st = 12 ((p q)2 + (q r)2 + (r p)2 ) > 0,
and the result follows. 

Now, we set p = a, q = b 3 2, and r = c 3 4, so by our lemma,


p3 + q 3 + r3 3pqr = a3 + 2b3 + 4c3 6abc

has the same sign as p + q + r =a + b 3


2 + c 3 4, provided that p = q = r does not hold. If
p = q = r does hold, then a = b 3 2 = c 3 4, which implies a = b = c = 0 because a, b, c are

USA Mathematical Talent Search


Solutions to Problem 3/4/16
www.usamts.org

integers. Thus, f (a, b, c) = a3 + 2b3 + 4c3 6abc = a + b + c = 0, and this case is covered as
well. This concludes our proof.
Solution 2 by: Laura Starkston (11/AZ)
If the signs must be the same, the zeros must be the same so...

3
3
a+b 2+c 4 = 0

3
3
a 2 + b 4 + 2c = 0

3
3
2 b 4 = 2c
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Keep this in mind. Rewrite the original equation:

3
3
a+b 2+c 4 =

3

3
a+b 2
=

3
3
a3 + 3 2a2 b + 3 4ab2 + 2b3 =
a3 + 2b3 + 4c3 =
a3 + 2b3 + 4c3 =

0


3

3
c 4

4c3

3
3
3 2a2 b 3 4ab2

3
3
(a 2 b 4)(3ab)

Combine the equations:


a3 + 2b3 + 4c3 = 6abc
a3 + 2b3 + 4c3 6abc = 0
Since the only operations performed were multiplication by a constant (which does not
change the sign or the zeros, only the magnitude of the values) and cubing (which does
not change the zeros; it makes each zero occur 3 times, but each zero is still the same; it
preserves the sign because it is an odd number power), f (a, b, c) where thefunction
is defined

3
3
3
3
3
as f (x, y, z) = x + 2y + 4z 6xyz should have the same sign as a + b 2 + c 2.

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