Algebra 431z+x=by.View this as a homogeneous system in the variablesx, y, z. Because we assume that
the system admits nonzero solutions, the determinant of the coefficient matrix is zero.
Writing down this fact, we obtain a new Diophantine equation in the unknownsa, b, c:
abc−a−b−c− 2 = 0.This can be solved by examining the following cases:
1.a=b=c. Thena=2 and it follows thatx=y=z, because these numbers are
pairwise coprime. This means thatx=y=z=1 andt=8. We have obtained the
solution( 1 , 1 , 1 , 8 ).2.a=b,a =c. The equation becomesa^2 c− 2 = 2 a+c, which is equivalent to
c(a^2 − 1 )= 2 (a+ 1 ), that is,c(a− 1 )=2. We either recover case 1, or find the
new solutionc=1,a=b=3. This yields the solution to the original equation
( 1 , 1 , 2 , 9 ).3.a>b>c. In this caseabc− 2 =a+b+c< 3 a. Therefore,a(bc− 3 )<2. It
follows thatbc− 3 <2, that is,bc <5. We have the following situations:
(i)b=2,c=1, soa=5 and we obtain the solution( 1 , 2 , 3 , 10 ).
(ii)b=3,c=1, soa=3 and we return to case 2.
(iii)b=4,c=1, so 3a= 7,which is impossible.In conclusion, we have obtained the solutions( 1 , 1 , 1 , 8 ),( 1 , 1 , 2 , 9 ),( 1 , 2 , 3 , 10 ),
and those obtained by permutations ofx, y, z.
(Romanian Mathematical Olympiad, 1995)
238.Note thatmcomparisons give rise to a homogeneous linear system ofmequations
withnunknowns, namely the masses, whose coefficients are− 1 , 0 ,and 1. Determining
whether all coins have equal mass is the same as being able to decide whether the solution
belongs to the one-dimensional subspace ofRnspanned by the vector( 1 , 1 ,..., 1 ). Since
the space of solutions has dimension at leastn−m, in order to force the solution to lie
in a one-dimensional space one needs at leastn−1 equations. This means that we need
to perform at leastn−1 comparisions.
(Mathematical Olympiad Summer Program, 2006)
239.We are given thata 0 =an+ 1 =0 andak− 1 − 2 ak+ak+ 1 =bk, withbk∈[− 1 , 1 ],
k= 1 , 2 ,...,n. Consider the linear system of equations
a 0 − 2 a 1 +a 2 =b 1 ,
a 1 − 2 a 2 +a 3 =b 2 ,