Three-Variable Homogeneous Cubic

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

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

Case 2: The challenge is to count number of solutions of first term. Lets use the substitution $X= 10 + x, Y=10+y, Z=10+z$ . and count number of solution of the equation $X+Y+Z=30$ where $0\leq X,Y,Z \leq 20$ . If we did not have any such constraint then number of solution by the stars and stripes would be: ${32 \choose 2} = 496$ . If one of the variables was over 21 then number of solution would be: 3* ${11 \choose 2} = 165$ 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: $496 - 165 = 331$

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

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

[Edits for clarity - Calvin]

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

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

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

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

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.

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.

Rearranging and factoring the given equation,

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

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 $x+y+z=0$ ,

$y+z=-x$

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

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

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

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

for $1\le x \le10$ 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.

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

we can divide into 2 case

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

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

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

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

in case 2 we get 331 triplets

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

Total of them is 331+21-1=351

for x+y+z=0, number = 6* (1+……+10) +1

( 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.

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

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

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

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