#### Question

Let (x,y,z) with x\geq y\geq z\geq 0 be integers such that

\dfrac{x^3+y^3+z^3}{3}=xyz+21

Find x.

Collected in the board: Algebraic proof questions

Steven Zheng posted 2 weeks ago

Multiplying both sides by 3, the statement is transformed to,

x^3+y^3+z^3 = 3xyz+63
(1)

which reminds us of the identity

x^3+y^3+z^3−3xyz = (x+y+z)(x^2+y^2+z^2−xy−yz−zx)

To derive the identity, let's start with the Cube of Binomial formula

(x+y)^3=x^3+y^3+3xy(x+y)

Convert it to the form of sum of two cubes

x^3+y^3 = (x+y)^3 - 3xy(x+y)

then plus z^3,

x^3+y^3 +z^3= (x+y)^3+z^3 - 3xy(x+y)

Apply the sum of two cudes identity on (x+y)^3 and z^3,

then,

x^3+y^3 +z^3= (x+y+z)^3- 3z(x+y)(x+y+z) - 3xy(x+y)

Subtracting both sides by 3xyz gives

x^3+y^3+z^3−3xyz = (x+y+z)^3- 3z(x+y)(x+y+z) - 3xy(x+y)-3xyz

x^3+y^3+z^3−3xyz = (x+y+z)^3- 3z(x+y)(x+y+z) - 3xy(x+y+z)

RHS =(x+y+z)[(x+y+z)^2-3zx-3yz-3xy]

Now we have proved the identity and (1) is transformed to,

(x+y+z)(x^2+y^2+z^2−xy−yz−zx) = 63

Apply the formula of square of trinomial on the 2nd factor, then

(x+y+z)[(x+y+z)^2 - 3xy−3yz−3zx] = 63
(2)

Let

A = x+y+z
(3)
B = xy+yz+zx
(4)

The statement (2) is simplified as

A(A^2-3B) = 63
(5)

\because \dfrac{x+y+z}{xy+yz+zx} = \dfrac{\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy} }{\dfrac{1}{z} +\dfrac{1}{x}+\dfrac{1}{y} } \leq 1

\therefore x+y+z \leq xy+yz+zx
(6)
\therefore A\leq B
(7)

Now let's find factors of 63

63 = 1\times 63 = 63 \times 1 = 3\times 21=21\times 3 = 7\times 9 = 9\times 7

1. When A = 1, A^2-3B = 63, B <0. Cancel this case.

2. When A = 63, A^2-3B = 1, B=\dfrac{62\times 64 }{3} , which is not integer. Cancel this case.

3. When A = 3, A^2-3B = 21, B = \dfrac{21-9}{3} =4. But these is no solution of positive integers for x,y,z . Can cel this case.

4. When A=21, A^2-3B = 3, B = \dfrac{21^2-3}{3} =146

Substituting the values of A and B to their definitions (3) and (4) gives the system of equations

\begin{cases} x+y+z = 21 \\ xy+yz+zx = 146 \end{cases}

We are not going to solve the system of equations coz it has more varibles than number of equations.

Using identity for square of difference of binomials, construct the following equations

(x-y)^2+(y-z)^2+(z-x)^2 = [(x+y+z)^2-6( xy+yz+zx)]

Substitute the value of A and B, we get

(x-y)^2+(y-z)^2+(z-x)^2 = 21^2 - 6\times 146 = 6
(8)

Since x\geq y\geq z\geq 0

There's only one possibility to make sure (8) is true.

\begin{cases} x-y=2 \\ y-z=1 \\ z-x =1 \end{cases}
(9)

Solve the system of equations of (9) gives

\begin{cases} x =8 \\ y=7 \\ z =6 \end{cases}

5, When A = 7, A^2-3B = 9, B is not integer. Cancel the case.

6. When A = 9, A^2-3B = 7, B is not integer. Cancel the case.

In Summary, there's only one solution for x,y,z when

\begin{cases} x =8 \\ y=7 \\ z =6 \end{cases}

such that

\dfrac{x^3+y^3+z^3}{3}=xyz+21

Steven Zheng posted 2 weeks ago

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