Find all triplets (x,y,z) that satisfy: x^2012+y^2012+z^2012=3, x^2013+y^2013+z^2013=3, x^2014+y^2014+z^2014=3 Trivial solution: (1,1,1)

Linda Peters 2022-09-23 Answered
Find all triplets ( x , y , z ) that satisfy:
x 2012 + y 2012 + z 2012 = 3 x 2013 + y 2013 + z 2013 = 3 x 2014 + y 2014 + z 2014 = 3
Trivial solution: ( 1 , 1 , 1 )
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Answers (2)

Amiya Watkins
Answered 2022-09-24 Author has 6 answers
Let e 1 , e 2 and e 3 be LHS of the the first, second and third equations respectivelly, then:
e 3 2 e 2 + e 1 = x 2012 ( x 2 2 x + 1 ) + y 2012 ( y 2 2 y + 1 ) + z 2012 ( z 2 2 z + 1 ) = 0 x 2012 ( x 1 ) 2 + y 2012 ( y 1 ) 2 + z 2012 ( z 1 ) 2 = 0
Since all the sumands are product of even powers, they cannot be negative, so they all are 1. That means that each unknown is either 0 or 1. But only if they all are 1 the original equations hold.
Generalize this to n equations of the same kind and n unknowns:
{ x 1 a + x 2 a + x 3 a + + x n a = n x 1 a + 1 + x 2 a + 1 + x 3 a + 1 + + x n a + 1 = n x 1 a + 2 + x 2 a + 2 + x 3 a + 2 + + x n a + 2 = n x 1 a + n 1 + x 2 a + n 1 + x 3 a + n 1 + + x n a + n 1 = n
With a even and n odd.
Then we wan define e j as the L H S of the j-th equation. Then:
k = 1 n ( n k ) ( 1 ) k + 1 e k = k = 1 n x k a ( x k 1 ) n 1 = 0
So same as before, each x k is either 1 or 0, but only the n-tuple ( 1 , 1 , , 1 ) satisfies the equation.
If all the RHS are m < n with m a positive integer, then the solutions are the n-tuples with m ones and n m zeros permutated.
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Ilnaus5
Answered 2022-09-25 Author has 2 answers
While the method already posted is nice, can't resist this hint - from the first and last equation, by the inequality of power means
x 2014 + y 2014 + z 2014 3 2014 x 2012 + y 2012 + z 2012 3 2012
with equality iff | x | = | y | = | z | = 1. Now using the second equation, the unique answer is obvious.
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