Kashyap's system of equations

For $z=0$ , easily we get 2 triples $(x,y,z)$ which are $(1,0,0)$ and $(0,1,0)$ .

For $z\neq0$ ,consider that $x^3+y^3=(x+y)(x^2-xy+y^2)$ which equivalent with $1-z^2=(1-z)(x^2-xy+y^2)$ Because $z\neq1$ , we get $x^2-xy+y^2=1+z$ and we know that $x+y=1-z$ . By adding these 2 equations, we get $x^2+y^2+x+y-xy=2$ which equivalent with $(x-y)^2+(x+1)^2+(y+1)^2=6$ Obvious that the triples $((x-y)^2,(x+1)^2,(y+1)^2)$ must be $(4,1,1)$ and its permutations. Easily we get 4 other triples which are $(-2,0,3);(0,-2,3);(-3,-2,6);(-2,-3,6)$ . Easy to verify that those 6 solutions satisfy the system of equations above. Therefore, there are $6$ solutions of triple $(x,y,z)$ .

We have $x+y\neq 0$ since $z\neq 1$ ; and $z=1-x-y$ . Thus, $x^2+y^2-xy=\frac{x^3+y^3}{x+y}=\frac{1-z^2}{1-z}=1+z=2-x-y\Rightarrow x^2-(y-1)x+y^2+y-2=0$

Now, consider the function $x^2-(y-1)x+y^2+y-2=0$ as a function of x. In order to have a solution, its discriminant $\Delta=(y-1)^2-4(y^2+y-2)=-3(y+3)(y-1)\geq 0$ . Since $y$ is an integer satisfying $-3(y+3)(y-1)\geq 0$ , so the possible values of $y$ are $-3;-2;-1;0;1$

If $y=1$ then $x=0$ and $z=0$

If $y=0$ then $x=1$ ; $z=0$ or $x=-2$ ; $z=3$

If $y=-1$ then $\Delta$ is not a square number. Thus, no integer value of $x$ exists.

If $y=-2$ then $x=-3$ ; $z=6$ or $x=0$ ; $z=3$

If $y=-3$ then $x=-2$ ; $z=6$

Together we have $6$ integer solutions $(x,y,z)$ available.

We have: x^3 + y^3 = 1 - z^2, so (x+y)(x^2+y^2-xy) = (1-z)(1+z) (1).
Because z \neq 1, so z-1 \neq 0 and x+y \neq 0.
Therefore, (1) becomes: x^2+y^2-xy = 1+z (2).
Because x+y = 1-z, so z = 1 - x - y.
Therefore, (2) becomes: x^2+y^2-xy = 2 - x - y.
Hence, 2x^2 + 2y^2 - 2xy + 2x + 2y = 4.
Hence, (x-y)^2 + (x+1)^2 + (y+1)^2 = 6.
Because there is only one way to write 6 as the sum of three square numbers: 6 = 1+1+4, we consider three cases:

Case 1: (x-y)^2 = 4, (x+1)^2 = 1, (y+1)^2 = 1.
Thus, x-y=2 or x-y=-2, x+1=1 or x+1=-1, y+1=1 or y+1=-1.
Hence, (x,y) = (0,-2) or (-2,0) and z=3

Case 2: (x-y)^2 = 1, (x+1)^2 = 4, (y+1)^2 = 1.
Thus, x-y=1 or -1, x+1=2 or -2, y+1=1 or -1.
Hence, (x,y) = (1,0) or (-3,-2). The values of z are 0 and 6 respectively.

Case 3: (x-y)^2 = 1, (x+1)^2 = 1, (y+1)^2 = 4.
This cases is similar to case 2, so we have 2 solutions: (x,y,z) = (0,1,0) or (-2,-3,6).

In conclusion, there are 6 solutions for (x,y,z): (0,-2,3), (-2,0,3), (1,0,0), (-3,-2,6), (0,1,0), (-2,-3,6).

We can write the 2nd equation as:

$(x+y)(x^2+y^2-xy)=(1+z)(1-z)$

Now, since $z=1-x-y$ and $z \neq 1$

We have:

$x^2+y^2-xy=1+1-x-y$

$x^2-xy+x+y^2+y-2=0$

$\Rightarrow x^2-x(y-1)+(y^2+y-2)=0$

Using the quadratic formula to solve for $x$ :

$x=\frac{(y-1) \pm (\sqrt{-3y^2-6y+9})}{2}$ (Simplified Form)

Now, we know that $x$ and $y$ are integers, first we need $\sqrt{-3y^2-6y+9}$ to be an integer.

Say, $\sqrt{-3y^2-6y+9}=k$ , where $k$ is an integer.

$\Rightarrow k^2=-3y^2-6y+9$

$3y^2+6y+(k^2-9)=0$

Solving for $y$ :

$y=\frac{-6 \pm \sqrt{144-12k^2}}{6}$

Since $y$ is an integer $144-12k^2 \geq 0$

$12k^2 \leq 144$

$k^2 \leq 12$

Using that $k$ is an integer $-3 \leq k \leq 3$

Now looking at $\frac{k^2}{3}=-y^2-2y+3$ , we can say that $k$ has to be a multiple of $3$ since $y$ itself is an integer and the RHS of the above equation will be an integer as well.

So we have $k=\pm 3$ , $k=0$

Now, we find values of $y$ based on these values of $k$ , if any integer values of $y$ exist for them:

Case 1: When $k=0$ , $y=\frac{-6 \pm 12}{6} \Rightarrow y=-3$ or $y=1$

Case 2: When $k=3$ , $y=\frac{-6 \pm 6}{6} \Rightarrow y=-2$ or $y=0$ . We wont check for $k=-3$ since it will give the same results as Case 2.

Now, we find $x$ .

$x=\frac{(y-1) \pm (\sqrt{-3y^2-6y+9})}{2}$

Note that $-(3y^2+6y-9)=-3(y-1)(y+3)$

So, for $y=1$ and $y=3$ we will have only $1$ integer value of $x$ for each. Using which we can get two values of $z$ respectively for $y=1$ and $y=-3$ .

For the other two values of $y$ there will be two integer values of $x$ each. So total number of solutions is $2+4=6$

Now, let us verify if they really exist or not.

Case 1: When $y=1$

We get $x=0$ for which we have $z=0$

So, $(0,1,0)$ is a solution.

Case 2: When $y=-3$

$x=-2$ and $z=6$

$(-2,-3,6)$ is a solution.

Case 3: When $y=-2$

$x=\frac{-3 \pm \sqrt{-3*-3*1}}{2}$

$\Rightarrow x=0$ or $x=-3$ which gives us $z=3$ and $z=6$ respectively.

Solutions: $(0,-2,3)$ and $(-3,-2,6)$

Case 4: When $y=0$

$x=\frac{-1 \pm 3}{2}$

$\Rightarrow x=1$ or $x=-2$ for which we have $z=0$ or $z=3$

So we have two more solutions $(1,0,0)$ and $(-2,0,3)$

Hence, we have verified that there are a total of $6$ solutions to the given system of equations.

$ Given x + y = 1 - z (1) ,

       x^3 + y^3 = 1 - z^2 (2)  $

divide (2) by (1) ,

$ x^{2} - xy + y^{2} = 1 + z ,

x^{2} - xy + y^{2} = 2 - x - y .

y^{2} - (x-1)*y + (x^{2} + x - 2) = 0 .(4) $

So , determinant of above equation D is $ \sqrt( - 3x^{2} - 6 x + 9) . $ "Necessary" condition :- D must be square of integer , if y is integer then , $ -3 x^{2} - 6*x + 9 = m^{2} ( where m is integer )

it is - 3 x^{2} - 6 x + (9 - m^{2}) = 0 . (4)

Determinant of above equation D" = \sqrt(144 - 12*m^{2}) => integer .

Taking values of m through 0 to 3 ( for m > 3 , D" will become complex) , for only m = 0 & m = 3, D" is integer . $

Put values of m in (4) and then values of x in (3) & solving them gives , six pairs of $ (x,y) : {(-3,-2), (-2,-3), (-2,0), (0,-2), (0,1), (1,0)} $ . for each pair of them there exist unique integer z (except 1) .

From the restriction, the first equation cannot equal zero.

Therefore, we can divide the second given equation by the first to obtain $x^2-xy+y^2=1+z$ .

The first equation implies $z=1-x-y$ , and substitution gives $x^2-xy+y^2=1+1-x-y$

$\Rightarrow y^2+(1-x)y+(x^2+x-2)=0$

For integer (and therefore real) $y$ we need the discriminant to be greater than or equal to $0$ : $(1-x)^2-4(1)(x^2-x-2)\geq0$ .

$-3(x^2+2x-3)\geq0$

$(x+1)^2-4\leq0 \Rightarrow -2\leq x+1\leq2$

Testing integer $x, -3\leq x \leq 1$ gives integer $y$ for only $(x,y) \in \{(-3, -2), (-2, -3), (-2, 0), (0, -2), (0, 1), (1, 0)\}$ .

Noting that the first given equation implies that $z$ is an integer if both $x$ and $y$ are, we are done with $\boxed{6}$ integer solutions $(x,y,z)$ .

Let $x+y = a$ and $x \times y = b$ .

$x^3 + y^3$ can be factored to $(x+y) \times (x^2 + y^2 - xy)$

so we have $a = 1-z$ as equation 1 and $(a)(a^2 - 3b)=1-z^2$ as equation 2.

$1-z^2 = (1-z)(1+z) = a(1+z)$

Equation 2 can now be simplified to $a^2 - 3b = 1+z$ . Let this be equation 3.

Plug in $a = 1-z$ for equation 3.

$1-2z+z^2 - 3b = 1+z$

$z^2 - 3z = 3b$

$\frac {z^3 - 3z} {3} = b$

$x$ and $y$ are the solutions to the equation $s^2 - as + b = 0 = s^2 - (1-z) s + \frac {z^3 - 3z} {3}$

Apply the quadratic formula:

$x$ and $y$ = $\frac{1-z \pm \sqrt{(1-z)^2-4(\frac {z^2 - 3z} {3})}}{2} =\frac{1-z \pm \sqrt{\frac {-z^2}{3} + 2z + 1}}{2}$

The maximum value of $\frac {-z^2}{3} + 2z + 1$ is 4 and occurs at $z = 3$ , so the possible values for $\sqrt {\frac {-z^2}{3} + 2z + 1}$ are 0,1,and 2.

$\sqrt {\frac {-z^2}{3} + 2z + 1} = 2$

$\frac {-z^2}{3} + 2z + 1 = 4$

$z = 3$

$x$ and $y$ = 0 and 2

$\sqrt {\frac {-z^2}{3} + 2z + 1} = 1$

$\frac {-z^2}{3} + 2z + 1 = 1$

$z = 0$

$x$ and $y$ = 0 and 1

$z = 6$

$x$ and $y$ = -2 and -3

$\sqrt {\frac {-z^2}{3} + 2z + 1} = 0$

$\frac {-z^2}{3} + 2z + 1 = 0$

$z= \frac{-2 \pm \frac{4 \sqrt{3}}{3}}{2}$

Since we are only looking for integer solutions, and $x$ and $y$ are interchangeable, there are $3 \times 2 = 6$ solutions.

Squaring the first equation, adding it to the second equation, and simplifying gives $(x+y)(x^2+y^2+x+y-xy-2)=0$ . So either $x+y=0$ , or $x^2+y^2+x+y-xy-2=0$ . But if $x+y=0$ , then $z=1$ , and we can't have that. Viewing the second part of the product as a quadratic in $x$ , and taking the discriminant gives $-3y^2-6y+9$ . This must be non-negative, because both $x$ and $y$ have to be integers. Factoring the discriminant gives $-3(y-1)(y+3)$ . $y$ must be between $-3$ and $1$ in order for the discriminant to be negative. From here, it is very easy to just test integer values of $y$ and get the corresponding values of $x$ and $z$ . Doing so gives $6$ unique solutions.

Substituting z from the first equation into the second: (x+y) (x^2-x y+x^2)=(x+y)(2-x-y)
Since z != 1, that is x+y != 0, contract both sides by (x+y):
x^2-x y+x^2+x+y=2, that is (x+1)^2-(x+1) (y+1)-(y+1)^2=3.
Let u=x+1, v=y+1. Then u^2-u v+v^2=3, or (u+v)^2+3 (u-v)^2=12.
Here abs(u+v) is bounded by sqrt(12), that is, by 3, and abs(u-v) is bounded by 2.
Noting that u+v and u-v should have the same evenness, we obtain that the only possible values for (u,v) are (-2,-1), (-1,-2), (-1, 1), (1, -1), (1, 2), (2, 1), all satisfying z != 1, that is u+v != 2.

Since $z\not=1$ , we can divide the second equation by the first equation to get $x^2-xy+y^2=1+z$ . Adding this new equation to $x+y=1-z$ gives us $x^2-xy+y^2+x+y=2$ . Observe that this is equivalent to $(x-y)^2+(x+1)^2+(y+1)^2=6$ .

Since $x,y$ are integers, the only way for three squares to sum to $6$ is $1+1+4=6$ . Thus, $x-y,x+1,y+1$ must be $\pm1,\pm1\pm2$ in some order. Noting that $x-y$ is determined from $x+1$ and $y+1$ , we can easily check the possibilities, and find that the following are the only ones that work:

$x+1=2$ and $y+1=1$ (and vice-versa)

$x+1=-1$ and $y+1=-2$ (and vice-versa)

$x+1=1$ and $y+1=-1$ (and vice-versa)

Hence, we have the $\boxed{6}$ solutions:

$(1,0,0); (0,1,0); (-2,-3,6); (-3,-2,6); (0,-2,3); (-2,0,3)$ .

Our equations are $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$ . The first equation can readily be written to $x = 1 - z - y$ . Cubing this, we obtain $x^3 = -z^3 - 3yz^2 + 3z^2 - 3y^2z + 6yz - 3z - y^3 + 3y^2 - 3y + 1$ .

Substituting this expression into the second equation and simplifying, we can obtain $-z^3 - 3yz^2 + 4z^2 - 3y^2z + 6yz - 3z + 3y^2 - 3y = 0$ . This is a quadratic in $y$ , $(-3z + 3)y^2 + (-3z^2 + 6z - 3)y + (-z^3 + 4z^2 - 3z)$ .

We inspect the discriminant of the quadratic. This is given by $(-3z^2 + 6z - 3)^2 - 4(-3z + 3)(-z^3 + 4z^2 - 3z))$ which can be simplified to $-3z^4 + 24z^3 - 30z^2 + 9$ .

We solve the roots of the discriminant in $z$ . By inspection we have $z = 1$ , and applying the factor theorem we can factor the quartic in $z$ to $-3(z-1)^2(z^2 - 6z - 3)$ . It thus follows that the other 2 roots are $z = 3 \pm 2\sqrt{3}$ using the quadratic formula. For values of $z$ smaller than $3 - 2\sqrt{3}$ or larger than $3 + 2\sqrt{3}$ , the discriminant is negative and thus there can be no real solutions and thus no integer solutions for $y$ .

We thus conclude that $z$ lies on the interval $[3 - 2\sqrt{3}, 3 + 2\sqrt{3}]$ and since we are only interested in the integer solutions of $z$ we need only consider $[0, 6]$ .

We then can use casework on the possible values of $z$ , substituting this into the quadratic of $y$ in terms of $z$ , and solving it. If $z = 0$ , then our quadratic for $y$ is $3y^2 - 3y = 0$ . This has the solutions $y = 0$ and $y = 1$ , which yield solutions of $x = 1$ and $x = 0$ respectively. Thus this value of $z$ gives us two solutions.

$z = 1$ is not allowed as we have our third equation $z \neq 1$ .

If $z = 2$ we have $-3y^2 - 3y + 2 = 0$ and the quadratic formula yields $y = \frac{1}{6}(-3 \pm \sqrt{33})$ - not an integer.

If $z = 3$ we have $-6y^2 - 12y = 0$ , which gives $y = -2$ and $y = 0$ , which respectively give $x = 0$ and $x = -2$ - two solutions.

If $z = 4$ we have $-9y^2 - 27y - 12 = 0$ which gives $y = \frac{1}{6}(-9 \pm \sqrt{33})$ - not an integer.

If $z = 5$ we have $-12y^2 - 48y - 40 = 0$ which gives $y = (-2 \pm \sqrt{\frac{2}{3}})$ - not an integer.

Finally if $z = 6$ we have $-15y^2 - 75y - 90 = 0$ which gives $y = -2$ and $y = -3$ , also yielding $x = -3$ and $x = -2$ respectively - two solutions.

Thus in total, we have two solutions each for the cases $z = 0, 3, 6$ and thus we have a total of $\boxed{6}$ solutions.

Dividing the second equation with the first equation we get

x ^2- xy + y ^2=1+ z

Adding this equation with the first equation we get

x ^2+ x - xy + y + y ^2=2

This is equivilant to

( x - y )^2+( x +1)( y +1)=3

Until this step, we split this equation into 4 cases.

a)When ( x - y )^2=0, x = y and thus ( x +1)^2=3 and there are no integer solutions.

b)When ( x - y )^2=1, ( x +1)( y +1)=2. Solving both equation we get ( x , y )=(0,1),(1,0),(-3,-2),(-2,-3)

c)When ( x - y )^2=2 or 3, there are no solutions

d)When ( x - y )^2=4, ( x +1)( y +1)=-1 After solving ti we got another 2 solutions, (1,-1) and (-1,1)

For the other numbers, there are no solutions because for the odd squares, the difference of the factors must be odd and for even perfect squares, the difference of the factors must be even.

Thus, the total number of solutions is 6.

x^3+y^3=(x+y)(x^2-xy+y^2) =(1-z)(x^2-xy+y^2) From second equation, (1-z)(x^2-xy+y^2)=(1+z)(1-z) (Since, 1-z^2=(1+z)(1-z)) So, x^2-xy+y^2=1+z From first equation, z=1-(x+y) So, x^2-xy+y^2+x+y-2=0.......3 Putting x=0 and then y=0, we get two quadratic equations which yield 4 solutions x^2+x-2=0 and y^2+y-2=0 Now, Equation 3 shall be arranged in the following way: y^2-y(x-1)+(x^2+x-2)=0; This is a quadratic equation with y as variable. Using the formula for roots of quadratic equation, we get y=((x-1)+-sqrt(9-6x-3x^2))/2 for y to be an integer, the term within the square root should be zero so 9-6x-3x^2 =0 which gives x=-3, and x=1 but x=1 has already been found as a solution. So, we have 5 solutions till now. Now, we can arrange equation 3 in a similar way, but this time making x the variable: x^2-x(y-1)+x^+x-2=0; Then, using a similar procedure as above, we find that there's one more solution. There can't be any more solutions, because the term within the square root of the quadratic roots formula cannot be a perfect square other than 1.

x+y = 1-z x^3 + y^3 = 1-z^2

(x+y)^2 = x^2 + 2xy + y^2 = (1-z)^2 (x^3+y^3) = (x+y)(x^2 - xy + y^2) = 1 - z^2 x^2 - xy + y^2 = 1+z (I divided by x+y, which equals 1-z) x^2+2xy+y^2 = 1 - 2z + z^2 (as written above) Subtract equations leaving... 3xy = z^2 - 3z = z(z-3)

All right, we have an equation and we know x,y,z must be integers. For this equation to be true, z must be a multiple of 3 or else x,y will not be integers.

If z is negative, [x+y = positive number] and [x^3 + y^3 = negative number] and [3xy = z^2 - 3z = positive number]. Let's think about this...to satisfy the first two, one of x or y is positive (but not both or else x^3 + y^3 will not be negative). If only 1 is positive though, 3xy = negative, which doesn't work in statement 3. Not all 3 can be true, therefore z cannot be negative.

If z = 0, x+y = 1 and x^3 + y^3 = 1 and 3xy = 0 x=1, y=0, z=0 x=0, y=1, z=0 are the only solutions here and relatively easy to find since one of x or y must be zero for 3xy = 0

If z is positive... z=3 3xy = 0, x+y = -2, x^3 + y^3 = -8 x=-2, y=0, z=3 x=0, y=-2, z=3 are both easy to find since x or y must be 0 to make 3xy = 0

If z = 6, 3xy =18, xy = 6, x+y = -5 (as a quick thinking point, realize here that from now on, x and y must both be negative, since x+y = 1-z = negative AND x^3 + y^3 = 1-z^2 AND 3xy = (z)(z-3). The last statement forces both x and y to be either positive or negative and the first two statements force x and y to be in the negative case.)

(-2,-3,6) (-3,-2,6) are solutions after a quick look at earlier equations.

Now...let's look at z=9 3xy = 54 xy = 18 x+y = -8 If x and y are both integers, their product is at a maximum when x=y. Max value of xy is 16 (x=y=-4). Thus, no solutions.

Now...let's look at z =12 xy = 108 x+y = -11 Max value of xy = 30 (x=-6 and y=-5)

You'll notice that xy is increasing at a rate too fast for x+y to keep up with, so there are no solutions after z = 6

Therefore, there are only 6 solutions. (1,0,0) (0,1,0) (0,-2,3) (-2, 0, 3) (-2,-3,6) (-3,-2,6)

Eliminating $z$ from the system of equations, we obtain

$x^3 + y^3 + (1 - (x + y))^2 = 1,$

$(x + y)(x^2 - xy + y^2 + x + y - 2) = 0.$

Since $x+y = 1-z \ne 0$ , it must be the case that

$x^2 - xy + y^2 + x + y - 2 = 0.$

This can be written as

$(2x - y + 1)^2 + 3(y + 1)^2 = 12.$

Since $x$ and $y$ are integral, the possibilities are:

$2x - y + 1 = 0, y + 1 = \pm 2 \mbox{ or } 2x - y + 1 = \pm 3, y + 1 = \pm1$

Analyzing all these cases, we obtain

$(x, y, z) = (0,1,0) , (-2, -3,6) , (1,0,0) , (0, -2,3) ,(-2,0,3) , (-3, -2,6).$

All of these cases satisfy $z \ne 1$ , hence the answer is $6$ .