Three-Variable Homogeneous Cubic

Algebra Level 5

x , y x, y and z z are integers such that 10 x , y , z 10 -10 \leq x, y, z \leq 10 . How many ordered triplets ( x , y , z ) (x, y, z) satisfy x 3 + y 3 + z 3 = 3 x y z x^3+y^3+z^3 = 3 xyz ?


The answer is 351.

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10 solutions

Firstly, we can factorize x 3 + y 3 + z 3 3 x y z x^3 + y^3 + z^3 - 3xyz into 1 2 ( x + y + z ) ( ( x y ) 2 + ( y z ) 2 + ( z x ) 2 ) \frac{1}{2}(x+y+z)( (x-y)^2 + (y-z)^2 + (z-x)^2) . So if, x 3 + y 3 + z 3 = 3 x y z x^3 + y^3 + z^3 = 3xyz then 1 2 ( x + y + z ) ( ( x y ) 2 + ( y z ) 2 + ( z x ) 2 ) = 0 \frac{1}{2}(x+y+z)( (x-y)^2 + (y-z)^2 + (z-x)^2) = 0 ( x + y + z ) = 0 \Rightarrow (x+y+z)=0 or ( x y ) 2 + ( y z ) 2 + ( z x ) 2 ) = 0 (x-y)^2 + (y-z)^2 + (z-x)^2) = 0 ( x + y + z ) = 0 \Rightarrow (x+y+z)=0 or x = y = z x=y=z

Case 1: The last condition will be valid for 21 values that is: 10 x = y = z 10 -10 \leq x=y=z \leq 10

Case 2: The challenge is to count number of solutions of first term. Lets use the substitution X = 10 + x , Y = 10 + y , Z = 10 + z X= 10 + x, Y=10+y, Z=10+z . and count number of solution of the equation X + Y + Z = 30 X+Y+Z=30 where 0 X , Y , Z 20 0\leq X,Y,Z \leq 20 . If we did not have any such constraint then number of solution by the stars and stripes would be: ( 32 2 ) = 496 {32 \choose 2} = 496 . If one of the variables was over 21 then number of solution would be: 3* ( 11 2 ) = 165 {11 \choose 2} = 165 Note, we cannot have more than one variable over 21, since the sum is only 30. Hence our desired count applying Inclusion-Exclusion principle is: 496 165 = 331 496 - 165 = 331

Both term will be valid for only 1 value that is: x = y = z = 0 x=y=z=0 .

Hence again applying inclusion exclusion principle, the total number of solution is: 331 + 21 1 = 351 331 + 21 - 1 = 351

[Edits for clarity - Calvin]

This problem was done quite badly. Without the factorization, students were unable to explain that there were only these 2 cases. Most complained that there were 21 solutions arising from x = y = z x=y=z . Students also had problems applying the Principle of Inclusion and Exclusion , and several students didn't account for the overlapped solution in Case 1 and Case 2.

Finally, Mahbubul shows a straightforward way to calculate solutions to x + y + z = 0 x+y + z = 0 under the restriction that 10 x , y , z 10 -10 \leq x, y, z \leq 10 . This method was used in a similar problem in the past.

Calvin Lin Staff - 7 years ago

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Sir this should be under combinatorics.

U Z - 6 years, 4 months ago

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Sometimes, you just can't categorize great problems.

Christopher Boo - 6 years, 4 months ago

Ya, I agee with you, and I think the problem I is little i er rated for having 24% solvers.

Trevor Arashiro - 6 years, 3 months ago
Avinash Pandey
May 20, 2014

We know that x 3 + y 3 + z 3 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 x y y z z x ) x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) So x 3 + y 3 + z 2 = 3 x y z x^3 + y^3 + z^2 = 3xyz is possible when:

i) x + y + z = 0 x+y+z=0 . In this case z = ( x + y ) z=-(x+y) . Hence z z is uniquely determined for each ordered pair ( x , y ) (x,y) . Since 10 z 10 -10 \leq z \leq 10 , we get 10 x + y 10 -10 \leq x+y \leq 10 . If we choose x = 10 x=-10 then we have 11 choices for y y i.e., 0 , 1 , 2... , 10 0,1,2...,10 . Similarly, for x = 9 x=-9 , we have 12 choices for y y i.e., 1 , 0 , 1 , 2... , 10 -1,0,1,2...,10 . The number of choices for y y goes on increasing by 1 1 from 11 11 ( when x = 10 ) x=-10) ) to 20 20 (when x = 1 x=-1 ). When 10 x 1 -10 \leq x \leq -1 , the number of ordered pairs ( x , y ) (x,y) is 11 + 12 + . . . + 20 = 155 11+12+...+20=155 . Same is the number of pairs when 1 x 10 1 \leq x \leq 10 and can be verified easily. When x = 0 x=0 , we have all the 21 21 choices for y y . So total number of pairs in this case is 2 155 + 21 = 331 2 \cdot 155 + 21 = 331 .

ii) x 2 + y 2 + z 2 x y y z z x = 0 x^2 + y^2 + z^2 - xy - yz - zx = 0 1 2 [ ( x y ) 2 + ( y z ) 2 + ( z x ) 2 ] = 0 \Rightarrow \frac{1}{2} [(x-y)^2 + (y-z)^2 + (z-x)^2] = 0 ( x y ) 2 + ( y z ) 2 + ( z x ) 2 = 0 \Rightarrow (x-y)^2 + (y-z)^2 + (z-x)^2 = 0 ( x y ) 2 = ( y z ) 2 = ( z x ) 2 = 0 \Rightarrow (x-y)^2 = (y-z)^2 = (z-x)^2 = 0 x y = y z = z x = 0 \Rightarrow x-y= y-z= z-x=0 x = y = z \Rightarrow x=y=z . In this case we are just looking for number of ways of selecting one element from set { 10 , 9 , 8... , 9 , 10 } \{ -10,-9,-8...,9,10 \} which is 21 21 . But we have also counted the pair ( 0 , 0 , 0 ) (0,0,0) which already has been counted in first case.

So, total number of ordered pairs ( x , y , z ) (x,y,z) is 331 + 21 1 = 351 331 + 21 - 1 = 351 .

Dingding Dong
May 20, 2014

According to the key technique, we can see there are: x+y+z=0 or x^2+y^2+z^2-xy-yz-xz=0. If x+y+z=0, we can count that there are 331 ordered triplets. If ^2+y^2+z^2-xy-yz-xz=0, there are 1/2[(x+y)^2+(y+z)^2+(x+z)^2]=0, x=y=z, so there are 10-(-10)+1=21 ordered triplets. Besides, (0,0,0) appears in both of these two situations and is counted twice. So the answer equals to 331+21-1=351.

Noel Lo
May 20, 2014

x^3 + y^3 +z^3 = 3xyz x^3 + y^3 +z^3 - 3xyz= 0 x^3 + (x^2)y +(x^2)z + y^3 +(y^2)x +(y^2)z + z^3 +(z^2)x +(z^2)y - (x^2)y - (y^2)x - xyz - (x^2)z - (z^2)x- xyz - (y^2)z- (z^2)y - xyz = 0 (x+y+z)(x^2 + y^2 + z^2 - xy- yz- xz) = 0 x+ y +z = 0 or x^2 + y^2 + z^2 - xy- yz- xz = 0 For the former solution, if x=0, y= -z and we have (y,z) = (10,-10), (9,-9) ....(-9,9), (-10,10) - a total of 21 solutions. For x modulo = 1 (i.e. x= +/- 1), we have 20 solutions for y and z. For x modulo 2, we have 19 solutions and so on. For x modulo 10, we have 11 solutions. Hence it can be seen that for x+y+z = 0, we have a total of 2(11+12+13+...+19+20)+ 21 = 331 solutions,

As for the latter solution as stated in the factorised form, we let x^2 = xy, y^2 = xz and z^2 = yz. The first equation tells us that x= y and the last one tells us that y=z, hence x=y=z. We therefore have (x,y,z) = (-10,-10,-10) all the way to (10,10,10) which gives us 21 solutions here. But do note that (0,0,0) is counted in this scenario as well as if x+y+z = 0.

Hence there are 331+21-1 = 351 solutions altogether.

Zi Song Yeoh
May 20, 2014

Rearranging and factoring the given equation,

1 2 ( x + y + z ) [ ( x y ) 2 + ( x z ) 2 + ( y z ) 2 ] = 0 \frac{1}{2}(x+y+z)[(x-y)^{2}+(x-z)^{2}+(y-z)^{2}]=0

If the latter is equal to 0,

x=y=z

We have 10-(-10)+1=21 solutions in total.

If the former is equal to 0,

Rearranging x + y + z = 0 x+y+z=0 ,

y + z = x y+z=-x

Claim : there are 21-|x| solutions for each 10 x 10 -10\le x\le10 .

Proof: for 10 x 0 -10\le x \le0 we have

(y,z) = (-x-10,10), (-x-9,9),...,(10.-x-10)

With x+21 = 21-|x| solutions.

for 1 x 10 1\le x \le10 we have

(y,z) = (10-x,-10),(9-x,-9),...,(-10,10-x)

With 21-x = 21-|x| solutions.

Therefore, the number of solutions in this case is

11+12+13+14+...+20+21+20+...+14+13+12+11=331

Note that only the triplet (0,0,0) appeared in both cases.

Therefore, the total number of triplets that satisfy the conditions is 331+21-1=351.

Kevin Sun
May 20, 2014

If x 3 + y 3 + z 3 3 x y z = 0 x^3+y^3+z^3-3xyz = 0 , then we factor this into ( x + y + z ) ( x 2 + y 2 + z 2 x y x z y z ) = 0 (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=0 . Either x + y + z = 0 x+y+z=0 or ( x y ) 2 + ( y z ) 2 + ( z x ) 2 = 0 (x-y)^2+(y-z)^2+(z-x)^2=0 , which is equivalent to x = y = z x=y=z . If x = y = z x=y=z , we obviously have 21 solutions. If x + y + z = 0 x+y+z=0 , we have a little casework to do. Since ( 0 , 0 , 0 ) (0,0,0) is already covered, we can do casework on how many 0's there are. \\ Case 1: One of x, y, and z are 0. Then WLOG x = 0 (multiply by 3). We then get y = z y=-z , so there are 20 possibilities, which gives a total of 60 in this case. \\ Case 2: None of x, y, and z are 0. In this case, we can WLOG x > 0, y < 0, and z < 0, so we need to multiply by 6 in the end. Then x = ( y ) + ( z ) x = (-y)+(-z) where x, -y, and -z are positive. This is just the number of ways to have 2 positive integers add up to less than or equal to 10. This is 10 choose 2 or 45. Thus this case gives 45*6 = 270. \\ In total, this gives 21 + 60 + 270 = 351 21+60+270= \fbox{351} ways.

we can divide into 2 case

case 1 if x=y=z so x 3 + y 3 + z 3 = 3 x y z x^{3}+y^{3}+z^{3}=3xyz that satisfy too all integer but because 10 x , y , z 10 -10\leq x,y,z\leq 10 so there is only 21 triplets

case 2 we know x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 ) ( x y + x z + y z ) ( x + y + z ) + 3 x y z x^{3}+y^{3}+z^{3}=(x+y+z)(x^{2}+y^{2}+z^{2})-(xy+xz+yz)(x+y+z)+3xyz if x + y + z = 0 x+y+z=0 we can conclude that x 3 + y 3 + z 3 = 3 x y z x^{3}+y^{3}+z^{3}=3xyz set: a = x + 10 a=x+10 b = y + 10 b=y+10 c = z + 10 c=z+10

we get 0 a , b , c 20 0\leq a,b,c\leq 20 the problems change become a + b + c = 30 a+b+c=30

use inklusi eksklusi to solve so we get, ( 32 2 ) 3 ( 11 2 ) = 331 \binom{32}{2}-3\binom{11}{2}=331

in case 2 we get 331 triplets

but in case 1 and case 2 we found same solution there is ( 0 , 0 , 0 ) (0,0,0)

Total of them is 331+21-1=351

黎 李
May 20, 2014

for x+y+z=0, number = 6* (1+……+10) +1

David Lin Kewei
May 20, 2014

( x^3+y^3+z^3 = 3xyz )\ ( \Leftrightarrow (x+y+z)( (x-y)^2 +(y-z)^2 + (z-x)^2) = 0 )\ Either the right bracket is 0 (i.e. ( x=y=z )\ ), or ( x+y+z )\ After a bit a calculation (which is too messy to type), the first case has 20 triples and the second has 331, which makes a total of 351.

Calvin Lin Staff
May 13, 2014

From the factorization, 0 = x 3 + y 3 + z 3 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 x y y z z x ) 0 = x^3 + y^3 + z^3 - 3 xyz = (x+y+z) ( x^2 + y^2 + z^2 - xy - yz - zx ) = 1 2 ( x + y + z ) ( ( x y ) 2 + ( y z ) 2 + ( z x ) 2 ) = \frac {1}{2} (x + y + z ) ( (x-y)^2 + (y-z)^2 + (z-x)^2 ) , we have the following cases:

Case 1: x = y = z x=y=z . There are 10 ( 10 ) + 1 = 21 10 - (-10) + 1 =21 such triplets.

Case 2: x + y + z = 0 x + y + z = 0 . For z = k z = k , we want x + y = k x+y = -k subject to 10 x , y 10 -10 \leq x, y \leq 10 . Consider the square grid where 10 x , y 10 -10 \leq x, y \leq 10 , and the line x + y = k x+y =-k intersecting. It follows that there are 21 k 21 - | k | possible pairs for ( x , y ) (x, y) . Now, taking the sum from k = 10 k=-10 to k = 10 k=10 , there are a total of 11 + 12 + 13 + + 21 + + 11 = 331 11+12+13+\ldots + 21 + \ldots + 11 = 331 such triplets.

The only triplet that is double counted (in both cases) is ( 0 , 0 , 0 ) (0,0,0) . Thus there are 21 + 331 1 = 351 21 +331 - 1 = 351 triplets that satisfy the equation.

Sir i cant understand what the square grid you are talking about

Rajat Bisht - 6 years, 4 months ago

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