Number theory- 2004 problem
Number Theory
Level
4
How many solutions are there for triplets which are integers
2
3
4
1
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1 solution
If we take one of the variables as 0, the other two go to 0 itself. Hence a solution is (0,0,0).
Now each of the three variables must be non-zero. RHS is divisible by 4.
Four cases arise:
all are odd
exactly 2 of x,y,z are odd
exactly 1 of x,y,z is odd
all are even
In case 1 and 3, LHS becomes odd which is not possible.
In case 2, LHS mod 4 gives 2 instead of 0.
Hence only possibility is case 4.
Now letting x = 2 * x1 , y = 2 * y1 , z = 2 * z1 gives
(x1)^2 + (y1)^2 + (z1)^2 = 4008 * x1 * y1 * z1
similar argument forces us to conclude that
x1 = 2 * x2 , y1 = 2 * y2 , z1 = 2 * z2 and hence
(x2)^2 + (y2)^2 + (z2)^2 = 8016 * x2 * y2 * z2
This goes on and hence no such x,y,z exists.