Diophantine Equality?
How many ordered solutions ( x , y , z ) of positive integers exist to the equation below?
( x + y + z + x y z ) 2 = ( x y + y z + z x ) 2 + 2 0 1 9
The answer is 15.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
A very good question. Keep it up.
Log in to reply
Thanks, but I actually didn't create it
Log in to reply
No problem. But from next time, try creating your own new and original problems.
Nice solution
Neat solution! You've missed a case, though: note that 2 0 1 9 = 3 × 6 7 3 , so there's a case to be eliminated before you can assume x + y + z + x y + y z + z x + x y z = 2 0 1 9 .
Log in to reply
Thank you. I thought since in that case, there is no solution, I shouldn't consider it.
Thanks! you always point out the invisible.
This can be factored as
(x + y + z + xy + yz + xz + xyz)(x + y + z - xy - yz - xz + xyz) = 2019
Now, x + y + z + xy + yz + xz + xyz = 2019
and (x + y + z - xy - yz - xz + xyz) = 1 as x, y and z are positive integers.
Or (1 + x)(1 + y)(1 + z) = 2020
and (1 - x)(1 - y)(1 - z) = 0
Solving above two equations, we get unordered solution sets as
(1, 1, 504) accounting 3 ordered sets
(1, 4, 201) accounting 6 ordered sets
(1, 100, 9) accounting 6 ordered sets
Therefore total ordered sets satisfying above equation is 15